Leetcode-Longest Substring Without Repeating Characters

题目：

Given a string, find the length of the longest substring without repeating characters.

Examples:

Given "abcabcbb", the answer is "abc", which the length is 3.

Given "bbbbb", the answer is "b", with the length of 1.

Given "pwwkew", the answer is "wke", with the length of 3. Note that the answer must be asubstring,"pwke" is a subsequence and not a substring.

题目要求，在给出的长字符串中，找出无重复的最长的子字符串，输出该子字符串的长度即可。思路比较简单，从头扫描长字符串，遇到重复即停下，记录子字符串长度。从下一位开始从新计数，重复前述步骤。两个点，一是判断重复，二是记录最大子字符串长度。第二点容易实现，设置变量max记录当前遇到的最大长度，设置变量cur记录当前子字符串长度，与max比较，决定是否更新max。第一点，最直观地，本想建立一个字符数组，记录当前子字符串出现过的字符，每次录入新字符就扫描查找是否重复。关键是，查找和判断（是否重复）。判断这一步骤肯定不能省下。能否在查找上省略呢？考虑到ASCII只能表示256个字符，建立一个大小为256的布尔数组，用于标记字符是否已经出现，视字符（ASCII码）为下标，便可直接找到对应字符，进而判断。

C++代码如下：

class Solution {public:    int lengthOfLongestSubstring(string s) {        int n = s.length();        int x = 0, y = 0;        int max = 0;        int cur = 0;        bool sign[256] = {false};//此处利用ASCII共能表示256个字母，省去查找的步骤，一步到位判断是否重复                 while(y < n) {        if(sign[s[y]]== false) {//无重复         sign[s[y]] = true;        y++;        }        else {//出现重复        while(sign[s[x]] != sign[s[y]]) {        sign[s[x]] = false;        x++;        }        sign[s[x]] = false;         x += 1;        }        cur = y - x;        max = max > cur ? max : cur;        }        return max;    }};