Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string. Example 1: Input:s1 = "ab" s2 = "eidbaooo"…
Given two strings s1 and s2, write a function to return true if s2 contains the permutation of s1. In other words, one of the first string's permutations is the substring of the second string. Example 1: Input:s1 = "ab" s2 = "eidbaooo"…
1.单纯的Unicode 转码 String a = "\u53ef\u4ee5\u6ce8\u518c"; a = new String(a.getBytes("UTF-16"),"Unicode"); 2.String 字符串中含有 Unicode 编码时,转为UTF-8 public static String decodeUnicode(String theString) { char aChar; int len = theString…
Given a string s and a list of strings dict, you need to add a closed pair of bold tag <b> and </b> to wrap the substrings in s that exist in dict. If two such substrings overlap, you need to wrap them together by only one pair of closed bold…
公众号:爱写bug(ID:icodebugs) 给定一个字符串,你需要反转字符串中每个单词的字符顺序,同时仍保留空格和单词的初始顺序. Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order. 示例 1: 输入: "Let's take LeetCode co…
统计字符串中的单词个数,这里的单词指的是连续的非空字符.请注意,你可以假定字符串里不包括任何不可打印的字符.示例:输入: "Hello, my name is John"输出: 5 详见:https://leetcode.com/problems/number-of-segments-in-a-string/description/ C++: 方法一: class Solution { public: int countSegments(string s) { int cnt=0; f…
1.1 Implement an algorithm to determine if a string has all unique characters. What if you cannot use additional data structure? 这道题让我们判断一个字符串中是否有重复的字符,要求不用特殊的数据结构,这里应该是指哈希表之类的不让用.像普通的整型数组应该还是能用的,这道题的小技巧就是用整型数组来代替哈希表,在之前Bitwise AND of Numbers Range 数…
统计字符串中的单词个数,这里的单词指的是连续的不是空格的字符. 请注意,你可以假定字符串里不包括任何不可打印的字符. 示例: 输入: "Hello, my name is John" 输出: 5 class Solution { public: int countSegments(string s) { int len = s.size(); string temp = ""; int res = 0; for(int i = 0; i < len; i++)…
