LeetCode算法第三题

最新推荐文章于 2025-08-07 16:42:30 发布

逐云者123

最新推荐文章于 2025-08-07 16:42:30 发布

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分类专栏： LeetCode与数据结构文章标签： leetcode 算法 substring

本文链接：https://blog.youkuaiyun.com/gaussrieman123/article/details/78500056

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本文探讨了如何在C语言中找到字符串中最长的无重复字符子串，并给出了一种解决方案。通过双层循环实现，文章最后提出挑战：如何进一步降低时间复杂度至O(n)。

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Description

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 a substring, “pwke” is a subsequence and not a substring.

Code

int isUnique(char* s){
    for(int i=0;i<strlen(s);i++){
        for(int j=i+1;j<strlen(s);j++){
            if(s[i]==s[j]){
                return -1;
                break;
            }
        }
    }
    return 1;
}

int lengthOfLongestSubstring(char* s) {
    int l = strlen(s);
    char *r = (char*)calloc(l,sizeof(char));
    int g_max = 0;
    for(int a=0; a<strlen(s); a++){
        r[a] = s[a];
        int b = 1;
        while(b<(l-a)){
            r[a+b]=s[a+b];
            if (1 == isUnique(r+a)){
                b ++;
            }
            else
                break;
        }
        g_max = (g_max > b)? g_max : b;
    }
    return g_max;
}

Conclusion

How to get the result in C with a time complexity of O(n) ?
Even we know that sliding window is the best way to traverse, but how to decrease the time complexity is still a problem.