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add Readme and solution file download by leetcode_generate
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‎001-two-sum/two-sum.py

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# -*- coding:utf-8 -*-
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# Given an array of integers, return indices of the two numbers such that they add up to a specific target.
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#
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# You may assume that each input would have exactly one solution.
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#
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#
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# Example:
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#
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# Given nums = [2, 7, 11, 15], target = 9,
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#
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# Because nums[0] + nums[1] = 2 + 7 = 9,
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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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# UPDATE (2016/2/13):
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# The return format had been changed to zero-based indices. Please read the above updated description carefully.
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class Solution(object):
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def twoSum(self, nums, target):
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"""
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:type nums: List[int]
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:type target: int
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:rtype: List[int]
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"""
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d = {}
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for i, v in enumerate(nums):
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if v in d:
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return [d[v],i]
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d[target-v] = i
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‎002-add-two-numbers/add-two-numbers.py

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# -*- coding:utf-8 -*-
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# You are given two linked lists representing two non-negative numbers. The digits are stored in reverse order and each of their nodes contain a single digit. Add the two numbers and return it as a linked list.
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#
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# Input: (2 -> 4 -> 3) + (5 -> 6 -> 4)
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# Output: 7 -> 0 -> 8
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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 addTwoNumbers(self, l1, l2):
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"""
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:type l1: ListNode
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:type l2: ListNode
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:rtype: ListNode
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"""
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if(l1 is None and l2 is None):
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return None
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head = ListNode(0)
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point = head
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carry = 0
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while l1 is not None and l2 is not None:
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s = carry + l1.val + l2.val
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point.next = ListNode(s % 10)
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carry = s / 10
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l1 = l1.next
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l2 = l2.next
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point = point.next
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while l1 is not None:
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s = carry + l1.val
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point.next = ListNode(s % 10)
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carry = s / 10
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l1 = l1.next
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point = point.next
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while l2 is not None:
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s = carry + l2.val
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point.next = ListNode(s % 10)
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carry = s / 10
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l2 = l2.next
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point = point.next
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if carry != 0:
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point.next = ListNode(carry)
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return head.next
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‎003-longest-substring-without-repeating-characters/longest-substring-without-repeating-characters.py

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# -*- coding:utf-8 -*-
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# Given a string, find the length of the longest substring without repeating characters.
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#
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# Examples:
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#
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# Given "abcabcbb", the answer is "abc", which the length is 3.
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#
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# Given "bbbbb", the answer is "b", with the length of 1.
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#
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# Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be a substring, "pwke" is a subsequence and not a substring.
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class Solution(object):
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def lengthOfLongestSubstring(self, s):
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"""
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:type s: str
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:rtype: int
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"""
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longest, start, visited = 0, 0, [False for _ in range(256)]
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for ind, val in enumerate(s):
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if not visited[ord(val)]:
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visited[ord(val)] = True
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else:
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while val != s[start]:
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visited[ord(s[start])] = False
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start += 1
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start += 1
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longest = max(longest, ind - start + 1)
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return longest
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‎004-median-of-two-sorted-arrays/median-of-two-sorted-arrays.py

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# -*- coding:utf-8 -*-
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# There are two sorted arrays nums1 and nums2 of size m and n respectively.
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#
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# Find the median of the two sorted arrays. The overall run time complexity should be O(log (m+n)).
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#
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# Example 1:
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#
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# nums1 = [1, 3]
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# nums2 = [2]
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#
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# The median is 2.0
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#
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#
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#
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# Example 2:
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#
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# nums1 = [1, 2]
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# nums2 = [3, 4]
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#
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# The median is (2 + 3)/2 = 2.5
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class Solution(object):
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def findMedianSortedArrays(self, nums1, nums2):
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"""
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:type nums1: List[int]
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:type nums2: List[int]
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:rtype: float
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"""
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nums = sorted(nums1 + nums2)
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t_len = len(nums)
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if t_len == 1:
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return nums[0]
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if t_len % 2:
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return nums[t_len/2]
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else:
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return (nums[t_len/2] + nums[t_len/2 -1]) /2.0

‎005-longest-palindromic-substring/longest-palindromic-substring.py

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# -*- coding:utf-8 -*-
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# Given a string s, find the longest palindromic substring in s. You may assume that the maximum length of s is 1000.
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#
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# Example:
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#
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# Input: "babad"
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#
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# Output: "bab"
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#
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# Note: "aba" is also a valid answer.
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#
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#
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#
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# Example:
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#
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# Input: "cbbd"
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#
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# Output: "bb"
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class Solution(object):
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def longestPalindrome(self, s):
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"""
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:type s: str
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:rtype: str
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"""
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longest, mid = "", (len(s) - 1) / 2
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i, j = mid, mid
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while i >= 0 and j < len(s):
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args = [(s, i, i), (s, i, i + 1), (s, j, j), (s, j, j + 1)]
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for arg in args:
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tmp = self.longestPalindromeByAxis(*arg)
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if len(tmp) > len(longest):
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longest = tmp
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if len(longest) >= i * 2:
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if len(longest) == 1:
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return s[0]
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return longest
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i, j = i - 1, j + 1
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return longest
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def longestPalindromeByAxis(self, s, left, right):
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while left >= 0 and right < len(s) and s[left] == s[right]:
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left, right = left - 1, right + 1
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return s[left + 1: right]

‎006-zigzag-conversion/zigzag-conversion.py

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# -*- coding:utf-8 -*-
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# The string "PAYPALISHIRING" is written in a zigzag pattern on a given number of rows like this: (you may want to display this pattern in a fixed font for better legibility)
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#
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# P A H N
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# A P L S I I G
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# Y I R
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#
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#
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# And then read line by line: "PAHNAPLSIIGYIR"
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#
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#
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# Write the code that will take a string and make this conversion given a number of rows:
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#
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# string convert(string text, int nRows);
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#
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# convert("PAYPALISHIRING", 3) should return "PAHNAPLSIIGYIR".
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class Solution(object):
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def convert(self, s, numRows):
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"""
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:type s: str
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:type numRows: int
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:rtype: str
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"""
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if not s or len(s) == 0 or numRows <= 0:
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return ""
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if numRows == 1:
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return s
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if len(s) % (numRows + numRows - 2):
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s = s + '#' * (numRows + numRows - 2 - (len(s) % (numRows + numRows - 2)))
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blocks = len(s)/(numRows + numRows - 2)
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res = ''
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for i in range(numRows):
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for j in range(blocks):
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if i == 0 or i == numRows-1:
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res += s[i + j*(numRows + numRows - 2)]
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else:
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res += s[i + j*(numRows + numRows - 2)]
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res += s[2*numRows-2-i + j*(numRows + numRows - 2)]
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return ''.join(res.split('#'))
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‎007-reverse-integer/reverse-integer.py

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# -*- coding:utf-8 -*-
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# Reverse digits of an integer.
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#
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#
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# Example1: x = 123, return 321
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# Example2: x = -123, return -321
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#
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#
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# click to show spoilers.
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#
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# Have you thought about this?
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#
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# Here are some good questions to ask before coding. Bonus points for you if you have already thought through this!
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#
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# If the integer's last digit is 0, what should the output be? ie, cases such as 10, 100.
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#
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# Did you notice that the reversed integer might overflow? Assume the input is a 32-bit integer, then the reverse of 1000000003 overflows. How should you handle such cases?
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#
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# For the purpose of this problem, assume that your function returns 0 when the reversed integer overflows.
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#
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#
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# Update (2014-11-10):
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# Test cases had been added to test the overflow behavior.
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class Solution(object):
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def reverse(self, x):
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"""
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:type x: int
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:rtype: int
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"""
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l = list(str(abs(x)))
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l.reverse()
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rst = int(''.join(l))
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if rst > 2147483647:
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return 0
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else:
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return rst if x>=0 else rst * (-1)
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‎008-string-to-integer-atoi/string-to-integer-atoi.py

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# -*- coding:utf-8 -*-
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# Implement atoi to convert a string to an integer.
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#
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# Hint: Carefully consider all possible input cases. If you want a challenge, please do not see below and ask yourself what are the possible input cases.
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#
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#
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# Notes:
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# It is intended for this problem to be specified vaguely (ie, no given input specs). You are responsible to gather all the input requirements up front.
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#
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#
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# Update (2015-02-10):
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# The signature of the C++ function had been updated. If you still see your function signature accepts a const char * argument, please click the reload button to reset your code definition.
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#
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#
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# spoilers alert... click to show requirements for atoi.
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#
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# Requirements for atoi:
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#
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# The function first discards as many whitespace characters as necessary until the first non-whitespace character is found. Then, starting from this character, takes an optional initial plus or minus sign followed by as many numerical digits as possible, and interprets them as a numerical value.
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#
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# The string can contain additional characters after those that form the integral number, which are ignored and have no effect on the behavior of this function.
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#
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# If the first sequence of non-whitespace characters in str is not a valid integral number, or if no such sequence exists because either str is empty or it contains only whitespace characters, no conversion is performed.
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#
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# If no valid conversion could be performed, a zero value is returned. If the correct value is out of the range of representable values, INT_MAX (2147483647) or INT_MIN (-2147483648) is returned.
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class Solution(object):
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def myAtoi(self, str):
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"""
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:type str: str
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:rtype: int
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"""
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import re
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p = re.compile(r'^[+-]?[0-9]+')
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m = re.match(p, str.strip())
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if m:
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ret = int(m.group())
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if ret < -0x80000000:
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return -0x80000000
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elif ret > 0x7fffffff:
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return 0x7fffffff
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return ret
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else:
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return 0
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‎009-palindrome-number/palindrome-number.py

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# -*- coding:utf-8 -*-
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# Determine whether an integer is a palindrome. Do this without extra space.
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#
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# click to show spoilers.
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#
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# Some hints:
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#
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# Could negative integers be palindromes? (ie, -1)
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#
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# If you are thinking of converting the integer to string, note the restriction of using extra space.
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#
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# You could also try reversing an integer. However, if you have solved the problem "Reverse Integer", you know that the reversed integer might overflow. How would you handle such case?
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#
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# There is a more generic way of solving this problem.
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class Solution(object):
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def isPalindrome(self, x):
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"""
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:type x: int
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:rtype: bool
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"""
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if x < 0:
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return False
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x = abs(x)
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l = len(str(x))
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i = 1
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while i < l / 2 + 1:
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head = (x / 10 ** (l-i)) % 10
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tail = (x % 10 ** i) if i == 1 else (x % 10 ** i) / (10 ** (i-1))
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if head != tail:
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return False
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i = i + 1
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return True

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