Leetcode经典试题：Longest Substring Without Repeating Characters解析

题目如下：

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.
https://leetcode.com/problems/longest-substring-without-repeating-characters/

我学习的优秀代码：

int lengthOfLongestSubstring(string s) {
        vector<int> dict(, -);
        int maxLen = , start = -;
        for (int i = ; i != s.length(); i++) {
            if (dict[s[i]] > start)
                start = dict[s[i]];
            dict[s[i]] = i;
            maxLen = max(maxLen, i - start);
        }
        return maxLen;
    }

来源：https://leetcode.com/problems/longest-substring-without-repeating-characters/discuss/1737/C%2B%2B-code-in-9-lines.

个人加英文注释版：

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        vector<int> dict(,-);//256 is the amount of ASCII and its expanding. use -1 to initialize,dict stores each letter's  index of their last position
        int length=,start=-;//initialize
        for(int i=;i<s.length();i++){// i is the index
            if(dict[s[i]]>start)// it means whether this letter is already contained in the choosed string
                start=dict[s[i]];// if true, the index of start should be reset to the index of the repeated word
            dict[s[i]]=i;//update the index of their last position
            length=max(length,i-start);//i and start both point at the same letter, so the real length is one letter less,i-start-1+1=i-start
        }
        return length;//return the largest length
    }
};

解析如下{

相关知识点{

ASCII码与拓展ASCII码{

ASCII 码使用指定的7 位或8 位二进制数组合来表示128种字符。另外，后128个称为扩展ASCII码。许多基于x86的系统都支持使用扩展（或“高”）ASCII。扩展ASCII 码允许将每个字符的第8 位用于确定附加的128 个特殊符号字符、外来语字母和图形符号。所以共计256个}；

C++ max与min函数的使用{

#include<algorithm>//引用头文件

min(a,b)或者max(a,b}会返回两个数中的较小数或者较大数，通常只用于两个数的大小比较}；

}；

内容解析{

这个算法最巧妙的地方是使用了哈希表，由于ASCII码的数量很小，只有256个，并且对应规则很明确（指的是字符和整数的对应规则），所以直接开了一个长度为256的数组做哈希表，用来保存这个字符上一次出现时的下标。由于哈希表的使用，查表的时间复杂度变成了O1级，使得速度的提升非常明显！这就是哈希表的力量hhhh

在if判断正确时，i和start指向的都是同样的字符，所选取的长度应该是要减掉一个的，但是由于尾减去头的结果会比字符个数少一。所以+1和-1相抵。

其他内容解析见这篇博文：https://www.cnblogs.com/ariel-dreamland/p/8668286.html}

}；