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class Solution(object): def lengthOfLongestSubstring(self, s): """ :type s: str :rtype: int """ repeatChar={} maxLen=0 left=0 right=0 for curChar in s: if repeatChar.has_key(curChar): maxLen = max(max

1. Java In-place solution: public class Solution { public String convert(String s, int numRows) { if(numRows=s.length()) return s; StringBuilder sb = new StringBuilder();

Implement strStr(). Returns the index of the first occurrence of needle in haystack, or -1 if needle is not part of haystack. Java public class Solution { public int strStr(String haystac

题目： Compare two version numbers version1 and version2. If version1 > version2 return 1, if version1 version2 return -1, otherwise return 0. You may assume that the version strings are non-empty

Rotate an array of n elements to the right by k steps. For example, with n = 7 and k = 3, the array [1,2,3,4,5,6,7] is rotated to [5,6,7,1,2,3,4]. Try to come up as many solutions as you can,

You are given a string, s, and a list of words, words, that are all of the same length. Find all starting indices of substring(s) in s that is a concatenation of each word in words exactly once and

Given an encoded string, return it's decoded string. The encoding rule is: k[encoded_string], where the encoded_string inside the square brackets is being repeated exactly k times. Note that k is