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;
}
};