LeetCode 3. Longest Substring Without Repeating Characters ( C ) 最短代码_given a string s, find the length of the longest s-优快云博客

本文链接：https://blog.youkuaiyun.com/JuzerTech/article/details/116034944

本文介绍了一种解决编程问题的方法，即在给定字符串中找到最长的没有重复字符的子字符串。通过遍历字符串并使用滑动窗口策略，更新子字符串的起始位置以确保无重复字符。示例代码展示了如何实现这一算法，包括处理不同输入情况。

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题目为给定一个字符串，寻找其中最长的子字符串，并且此子字符串不得有重复之字符。

题目与范例如下

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

Example 1:

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

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

Input: s = "pwwkew"
Output: 3
Explanation: The answer is "wke", with the length of 3.
Notice that the answer must be a substring, "pwke" is a subsequence and not a substring.
Example 4:

Input: s = ""
Output: 0

Constraints:

0 <= s.length <= 5 * 104
s consists of English letters, digits, symbols and spaces

我采用的策略为，使用一个变量纪录子字符串的起始点，子字符串的终止点为目前寻访到的字符，并逐次检查新加入的字符是否与子字符串内的字符有重复，有则把子字符串的起始点设为重复的下一个，每次比较是否比记录中的最长长度还长。

下方为我的代码

void comparechar(char *str,char target,int *submin, int *submax){
    for(int i = *submin ; i < *submax ; i++ ){
        if(str[i] == target){
            *submin = i+1;
            break;
        }
    }
}

int lengthOfLongestSubstring(char * s){
    int index = 0,submin = 0,max = 0;
    while(s[index] != '\0'){
        comparechar(s,s[index],&submin,&index);
        if(index - submin + 1 > max)
            max = index - submin + 1;
        index++;
    }
    return max;
}