8000 update at 2019-07-25 · Frantol910/leetcode@a51c4a6 · GitHub
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update at 2019-07-25
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‎README.md

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‎solutions/0001-two-sum/two-sum.py

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# return [0, 1].
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#
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#
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#  
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#
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class Solution(object):

‎solutions/0001-two-sum/two-sum.rs

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// return [0, 1].
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//
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//
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//  
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//
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use std::collections::HashMap;

‎solutions/0003-longest-substring-without-repeating-characters/longest-substring-without-repeating-characters.py

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#
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#
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class Solution(object):

‎solutions/0010-regular-expression-matching/regular-expression-matching.py

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# -*- coding:utf-8 -*-
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# Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.
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# Given an input string (s) and a pattern (p), implement regular expression matching with support for '.' and '*'.
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#
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#
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# '.' Matches any single character.
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# '*' Matches zero or more of the preceding element.
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# '.' Matches any single character.
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# '*' Matches zero or more of the preceding element.
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#
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#
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# The matching should cover the entire input string (not partial).
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# The matching should cover the entire input string (not partial).
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#
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# Note:
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# Note:
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#
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#
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# s could be empty and contains only lowercase letters a-z.
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# p could be empty and contains only lowercase letters a-z, and characters like . or *.
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# s could be empty and contains only lowercase letters a-z.
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# p could be empty and contains only lowercase letters a-z, and characters like . or *.
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#
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#
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# Example 1:
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# Example 1:
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#
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#
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# Input:
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# s = "aa"
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# p = "a"
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# Output: false
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# Explanation: "a" does not match the entire string "aa".
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# Input:
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# s = "aa"
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# p = "a"
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# Output: false
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# Explanation: "a" does not match the entire string "aa".
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#
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#
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# Example 2:
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# Example 2:
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#
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#
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# Input:
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# s = "aa"
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# p = "a*"
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# Output: true
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# Explanation: '*' means zero or more of the precedeng element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
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# Input:
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# s = "aa"
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# p = "a*"
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# Output: true
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# Explanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".
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#
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#
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# Example 3:
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# Example 3:
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#
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#
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# Input:
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# s = "ab"
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# p = ".*"
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# Output: true
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# Explanation: ".*" means "zero or more (*) of any character (.)".
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# Input:
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# s = "ab"
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# p = ".*"
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# Output: true
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# Explanation: ".*" means "zero or more (*) of any character (.)".
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#
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#
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# Example 4:
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# Example 4:
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#
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# Input:
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# s = "aab"
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# p = "c*a*b"
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# Output: true
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# Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches "aab".
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# Input:
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# s = "aab"
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# p = "c*a*b"
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# Output: true
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# Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore, it matches "aab".
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#
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# Example 5:
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# Example 5:
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#
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# Input:
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# s = "mississippi"
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# p = "mis*is*p*."
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# Output: false
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# Input:
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# s = "mississippi"
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# p = "mis*is*p*."
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# Output: false
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‎solutions/0056-merge-intervals/merge-intervals.py

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# Output: [[1,5]]
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# Explanation: Intervals [1,4] and [4,5] are considered overlapping.
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#
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# NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
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#
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# Definition for an interval.

‎solutions/0057-insert-interval/insert-interval.py

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# Output: [[1,2],[3,10],[12,16]]
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# Explanation: Because the new interval [4,8] overlaps with [3,5],[6,7],[8,10].
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#
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# NOTE: input types have been changed on April 15, 2019. Please reset to default code definition to get new method signature.
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#
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# Definition for an interval.

‎solutions/0058-length-of-last-word/length-of-last-word.py

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# Example:
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#
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# Input: "Hello World"
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# Output: 5
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#  
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class Solution(object):

‎solutions/0101-symmetric-tree/symmetric-tree.py

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# Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
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# For example, this binary tree [1,2,2,3,4,4,3] is symmetric:
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# 1
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# / \
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# 2 2
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# / \ / \
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# 3 4 4 3
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#  
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# But the following [1,2,2,null,3,null,3] is not:
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# But the following [1,2,2,null,3,null,3] is not:
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# 3 3
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# Note:
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# Bonus points if you could solve it both recursively and iteratively.

‎solutions/0141-linked-list-cycle/linked-list-cycle.py

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# -*- coding:utf-8 -*-
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# Given a linked list, determine if it has a cycle in it.
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#
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# To represent a cycle in the given linked list, we use an integer pos which represents the position (0-indexed) in the linked list where tail connects to. If pos is -1, then there is no cycle in the linked list.
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#
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# Example 1:
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#
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#
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# Input: head = [3,2,0,-4], pos = 1
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# Output: true
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# Explanation: There is a cycle in the linked list, where tail connects to the second node.
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#
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# Example 2:
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# Input: head = [1,2], pos = 0
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# Output: true
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# Explanation: There is a cycle in the linked list, where tail connects to the first node.
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# Example 3:
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# Input: head = [1], pos = -1
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# Output: false
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# Explanation: There is no cycle in the linked list.
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# Follow up:
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# Can you solve it using O(1) (i.e. constant) memory?
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#
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# Definition for singly-linked list.
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# class ListNode(object):
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# def __init__(self, x):
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# self.val = x
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# self.next = None
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class Solution(object):
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def hasCycle(self, head):
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"""
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:type head: ListNode
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:rtype: bool
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"""
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if head is None:
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return False
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cur = head
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dct = {}
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while cur.next:
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if id(cur) in dct:
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return True
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dct[id(cur)] = cur
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cur = cur.next
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return False
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