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…

Given a string, find the first non-repeating character in it and return it's index. If it doesn't exist, return -1. Examples: s = "leetcode" return 0. s = "loveleetcode", return 2. Note: You may assume the string contain only lowercase…

题目标签:String, HashMap 题目给了我们一个 string,让我们找出 第一个 唯一的 char. 设立一个 hashmap,把 char 当作 key,char 的index 当作value. 遍历string,如果这个 char 没有在 map 里,说明第一次出现,存入 char,index:                    如果这个 char 已经在 map 里,说明不是第一次出现,而且我们不在乎这种情况,更新 char, -1. 遍历hashmap,把最小的 inde…

给定一个字符串,找到它的第一个不重复的字符,并返回它的索引.如果不存在,则返回 -1.案例:s = "leetcode"返回 0.s = "loveleetcode",返回 2.注意事项:您可以假定该字符串只包含小写字母.详见:https://leetcode.com/problems/first-unique-character-in-a-string/description/ C++: class Solution { public: int firstUniq…

给定一个字符串,找到它的第一个不重复的字符,并返回它的索引.如果不存在,则返回 -1. 案例: s = "leetcode" 返回 0. s = "loveleetcode", 返回 2. 注意事项:您可以假定该字符串只包含小写字母. class Solution { public: int firstUniqChar(string s) { int len = s.size(); map<char, int> check; for(int i = 0;…

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…

/** * 方法名称:replaceBlank * 方法描述: 将string字符串中的换行符进行替换为"" * */ public static String replaceBlank(String str) { String dest = ""; if (str != null) { Pattern p = Pattern.compile("\t|\r|\n"); Matcher m = p.matcher(str); dest = m.re…

Given a set of keywords words and a string S, make all appearances of all keywords in S bold. Any letters between <b> and </b> tags become bold. The returned string should use the least number of tags possible, and of course the tags should fo…

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…

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"…