题目地址：Longest Substring Without Repeating Characters

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题目简介：给定一个字符串，找到最长的无重复字符串

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

Example 1:

Input: "abcabcbb"
Output: 3 
Explanation: The answer is "abc", with the length of 3. 
Example 2:

Input: "bbbbb"
Output: 1
Explanation: The answer is "b", with the length of 1.
Example 3:

Input: "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3. 
Note that the answer must be a substring, "pwke" is a subsequence and not a substring.

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题目思路：

根据人为的寻找习惯，应该从左到右以此遍历字符串，在出现重复字符串时重新寻找子字符串。字符串中对应的ASCII码中的256个字符，为了方便记下这256个字符可能出现的位置，所以可以用一个数据结构存储他们的位置。

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C++版：

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
    int v[256]; //用来存储256个字符的位置
    memset(v,-1,sizeof(v)); //刚开始所有的位置均未初始化
    int len = s.length(), ans = 0,start = -1;//获得字符串的长度，ans代表最终的结果,start代表
                                                //子字符串的其实位置
    if (len <= 1) //如果输入的字符串为空或只有一个字符
        return len;
    for(int i = 0;i < len;i++)
    {
        if(v[s[i] - ' '] > start) //若下一个可以加入的字符在之前出现过，则更新子字符串的起始位置
            start = v[s[i] - ' '];
        v[s[i] - ' '] = i; //当前子字符串能到达的位置
        ans = max(ans,i - start);//更新最终的结果
    }
    return ans;
    }
};

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Python版：

class Solution:
    def lengthOfLongestSubstring(self, s: str) -> int:
        length = len(s) #获得字符串长度
        if length <= 1:
            return length
        start = -1
        ans = 0
        list1 = [-1] * 256 #列表初始化为256个-1
        for i in range(0,length):
            if list1[ord(s[i]) - ord(' ')] > start:
                start = list1[ord(s[i]) - ord(' ')];
            list1[ord(s[i]) - ord(' ')] = i
            ans = max(ans, i - start)
        return ans