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key points
Signed-off-by: Qin Yafei <yafee.qin@gmail.com>
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‎Keypoints.md

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# 1.线性表
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## 数组
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Remove Duplicates from Sorted Array
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||Remove Duplicates from Sorted Array II
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||Longest Consecutive Sequence
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||Two Sum
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||3Sum
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||3Sum Closest
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||4Sum
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||Remove Element
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||Move Zeroes
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||Next Permutation
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||Permutation Sequence
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||Valid Sudoku
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||Trapping Rain Water
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||Rotate Image
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||Plus One
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||Climbing Stairs
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||Set Matrix Zeroes
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||Gas Station
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||Candy
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||Majority Element
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||Rotate Array
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||Contains Duplicate
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||Contains Duplicate II
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||Contains Duplicate III
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||Product of Array Except Self
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||Game of Life
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||Increasing Triplet Subsequence
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## 单链表
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Reverse Linked List
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||Odd Even Linked List
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||Add Two Numbers
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||Reverse Linked List II
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||Partition List
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||Remove Duplicates from Sorted List
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||Remove Duplicates from Sorted List II
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||Rotate List
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||Remove Nth Node From End of List
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||Swap Nodes in Pairs
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||Reverse Nodes in k-Group
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||Copy List with Random Pointer
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||Linked List Cycle
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||Linked List Cycle II
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||Reorder List
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||LRU Cache
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||Palindrome Linked List
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# 2.字符串
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Valid Palindrome
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||Implement strStr
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||String to Integer (atoi)
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||Add Binary
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||**Longest Palindromic Substring**
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||Regular Expression Matching
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||Wildcard Matching
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||Longest Common Prefix
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||Valid Number
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||Integer to Roman
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||Roman to Integer
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||Count and Say
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||Anagrams
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||Valid Anagram
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||Simplify Path
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||Length of Last Word
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||Isomorphic Strings
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||Word Pattern
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# 3.栈和队列
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## 栈
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Min Stack
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||Valid Parentheses
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||Longest Valid Parentheses
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||Largest Rectangle in Histogram
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||Evaluate Reverse Polish Notation
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||Implement Stack using Queues
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## 队列
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||Implement Queue using Stacks
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# 4.二叉树
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## 二叉树的遍历
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Binary Tree Preorder Traversal
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||Binary Tree Inorder Traversal
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||Binary Tree Postorder Traversal
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||Binary Tree Level Order Traversal
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||Binary Tree Level Order Traversal II
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||Binary Tree Right Side View
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||Invert Binary Tree
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||Binary Search Tree Iterator
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||Binary Tree Zigzag Level Order Traversal
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||Recover Binary Search Tree
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||Same Tree
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||Symmetric Tree
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||Balanced Binary Tree
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||Flatten Binary Tree to Linked List
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||Populating Next Right Pointers in Each Node II
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## 二叉树的构建
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||Construct Binary Tree from Preorder and Inorder Traversal
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||Construct Binary Tree from Inorder and Postorder Traversal
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## 二叉查找树
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Unique Binary Search Trees
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||Unique Binary Search Trees II
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||Validate Binary Search Tree
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||Convert Sorted Array to Binary Search Tree
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||Convert Sorted List to Binary Search Tree
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||LCA of BST
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||Kth Smallest Element in a BST
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## 二叉树的递归
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Minimum Depth of Binary Tree
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||Maximum Depth of Binary Tree
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||Path Sum
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||Path Sum II
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||Binary Tree Maximum Path Sum
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||Populating Next Right Pointers in Each Node
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||Sum Root to Leaf Numbers
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||LCA of Binary Tree
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## 线段树
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||Range Sum Query - Mutable
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# 5.排序
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## 插入排序
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Insertion Sort List
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## 归并排序
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Merge Two Sorted Arrays
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Merge Two Sorted Lists
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Merge k Sorted Lists
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Sort List
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## 快速排序
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Sort Colors
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Kth Largest Element in an Array
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## 桶排序
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First Missing Positive
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## 计数排序
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H-Index
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## 基数排序
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Maximum Gap
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## 其他
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Largest Number
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# 6.查找
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Search for a Range
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||Search Insert Position
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||Search in Rotated Sorted Array
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||Search in Rotated Sorted Array II
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||Search a 2D Matrix
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||Search a 2D Matrix II
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||Find Minimum in Rotated Sorted Array
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||Find Minimum in Rotated Sorted Array II
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||Median of Two Sorted Arrays
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||H-Index II
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# 7.暴力枚举法
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Subsets
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||Subsets II
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||Permutations
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||Permutations II
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||Combinations
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||Letter Combinations of a Phone Number
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||# 8.广度优先搜索
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||Word Ladder
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||Word Ladder II
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||Surrounded Regions
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# 9.深度优先搜索
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Additive Number
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||Palindrome Partitioning
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||Unique Paths
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||Unique Paths II
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||N-Queens
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||N-Queens II
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||Restore IP Addresses
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||Combination Sum
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||Combination Sum II
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||Combination Sum III
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||Generate Parentheses
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||Sudoku Solver
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||Word Search
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# 10.分治法
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||Pow(x,n)
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||Sqrt(x)
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# 11.贪心法
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Jump Game
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||Jump Game II
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||Best Time to Buy and Sell Stock
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||Best Time to Buy and Sell Stock II
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||Longest Substring Without Repeating Characters
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||Container With Most Water
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||Patching Array
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# 12.动态规划
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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|120|Triangle|
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||Maximum Subarray
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||Maximum Product Subarray
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||Longest Increasing Subsequence
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||Palindrome Partitioning II
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||Maximal Rectangle
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||Best Time to Buy and Sell Stock III
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||Best Time to Buy and Sell Stock IV
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||Best Time to Buy and Sell Stock with Cooldown
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||Interleaving String
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||Scramble String
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||Minimum Path Sum
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||Edit Distance
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||Decode Ways
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||Distinct Subsequences
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||Word Break
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||Word Break II
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||Dungeon Game
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||House Robber
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||House Robber II
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||House Robber III
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||Range Sum Query - Immutable
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||Range Sum Query 2D - Immutable
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# 13.图
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||Clone Graph
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# 14.位操作
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Reverse Bits
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||Repeated DNA Sequences
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||Number of 1 Bits
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||Gray Code
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||Single Number
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||Single Number II
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||Single Number III
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||Power of Two
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||Missing Number
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||Maximum Product of Word Lengths
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||Bitwise AND of Numbers Range
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||Power of Three
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||Rectangle Area
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# 15.数论
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Happy Number
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||Ugly Number
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||Ugly Number II
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||Super Ugly Number
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||Fraction to Recurring Decimal
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||Factorial Trailing Zeroes
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||Nim Game
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# 16.模拟
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| # | Title | Difficulty | Solution |
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|---| ----- | -------- | ---------- |
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||Reverse Integer
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||Palindrome Number
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||Insert Interval
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||Merge Intervals
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||Minimum Window Substring
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||Multiply Strings
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||Substring with Concatenation of All Words
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||Pascal's Triangle
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||Pascal's Triangle II
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||Spiral Matrix
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||Spiral Matrix II
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||ZigZag Conversion
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||Divide Two Integers
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||Text Justification
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||Max Points on a Line

‎algorithms/cpp/3Sum/3Sum.cpp

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@@ -40,21 +40,21 @@ class Solution {
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sort(nums.begin(), nums.end());
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for (int i = 0; i < n - 2; ++i) {
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if (nums[i] > 0) { //已排序，后面不会有解
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if (nums[i] > 0) { // 已排序，后面不会有解
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return res;
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}
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if (i > 0 && nums[i] == nums[i-1]) { // 去重复
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if (i > 0 && nums[i] == nums[i - 1]) { //! 去重复
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continue;
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}
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auto left = i + 1;
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auto right = n - 1;
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while (left < right) {
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if (nums[i] + nums[left] + nums[right] == 0) {
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res.push_back({nums[i], nums[left], nums[right]});
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++left; //找下一个解
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--right; //找下一个解
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while (nums[left] == nums[left-1] && left < right) ++left; //去重复
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while (nums[right] == nums[right + 1] && left < right) --right; //去重复
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++left; //! 找下一个解
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--right; //! 找下一个解
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while (nums[left] == nums[left - 1] && left < right) ++left; // 去重复
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while (nums[right] == nums[right + 1] && left < right) --right; // 去重复
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} else if (nums[i] + nums[left] + nums[right] < 0) {
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++left;
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sort(nums.begin(), nums.end());
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auto last = nums.end();
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for (auto i = nums.begin(); i < last - 2; ++i) {
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if (*i > 0) return res; //已排序，后面不会有解
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if (i > nums.begin() && *i == *(i - 1)) continue; //去重复
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if (*i > 0) return res; // 已排序，后面不会有解
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if (i > nums.begin() && *i == *(i - 1)) continue; // 去重复
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auto left = i + 1;
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auto right = last - 1;
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while (left < right) {
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if (*i + *left + *right == 0) {
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res.push_back({*i, *left, *right});
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++left; //找下一个解
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--right; //找下一个解
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while (*left == *(left - 1) && left < right) ++left; //去重复
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while (*right == *(right + 1) && left < right) --right; //去重复
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++left; // 找下一个解
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--right; // 找下一个解
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while (*left == *(left - 1) && left < right) ++left; // 去重复
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while (*right == *(right + 1) && left < right) --right; // 去重复
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} else if (*i + *left + *right < 0) {
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++left;
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};
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#include <stdio.h>
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#include <algorithm>
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#include <iostream>
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#include <set>

‎algorithms/cpp/3SumClosest/3SumClosest.cpp

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@@ -26,10 +26,10 @@ class Solution {
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const int n = nums.size();
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int res = 0;
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int min_gap = INT_MAX;
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sort(nums.begin(), nums.end()); //先排序
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sort(nums.begin(), nums.end()); // 先排序
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for (int i = 0; i < n - 2; ++i) { // i<倒数第二个
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int left = i+1;
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int right = n-1;
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int left = i + 1;
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int right = n - 1;
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while (left < right) {
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const int sum = nums[i] + nums[left] + nums[right];
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const int gap = abs(sum - target);
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// if(nums.size() < 3) return;
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int res = 0;
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int min_gap = INT_MAX;
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sort(nums.begin(), nums.end()); //先排序
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sort(nums.begin(), nums.end()); // 先排序
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for (auto i = nums.begin(); i < prev(nums.end(), 2); ++i) { // i<倒数第二个
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auto left = next(i);
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auto right = prev(nums.end());
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#include <stdio.h>
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#include <stdlib.h>
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#include <algorithm>
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#include <iostream>
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#include <set>

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