[LeetCode] Longest Substring Without Repeating Characters

最新推荐文章于 2025-08-12 17:21:36 发布

转载最新推荐文章于 2025-08-12 17:21:36 发布 · 74 阅读

文章标签：

#c/c++

本文介绍了一种求解字符串中最长无重复字符子串长度的高效算法，并提供了详细的C++实现代码。通过维护字符最后出现的位置和当前有效子串的起始位置，算法能在O(n)时间内解决该问题。

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Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without repeating letters for "abcabcbb" is "abc", which the length is 3. For "bbbbb" the longest substring is "b", with the length of 1.

http://blog.youkuaiyun.com/pickless/article/details/9018575

#if 0
class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        int locs[256];//保存字符上一次出现的位置
        memset(locs, -1, sizeof(locs));

        int idx = -1, max = 0;//idx为当前子串的开始位置-1
        for (int i = 0; i < s.size(); i++)
        {
            if (locs[s[i]] > idx)//如果当前字符出现过，那么当前子串的起始位置为这个字符上一次出现的位置+1
            {
                idx = locs[s[i]];
            }

            if (i - idx > max)
            {
                max = i - idx;
            }

            locs[s[i]] = i;
        }
        return max;
    }
};
#endif
class Solution 
{
    public:   
        int lengthOfLongestSubstring(string s)  
        {   
            int hash[256]; // save the index of s[i]
            int  start = -1; 
            int len = 0;

            memset(hash, -1, sizeof(hash));

            for(int i = 0; i< s.size(); i++/**/)
            {   
                // update the start if the start is in the front of the old hash[s[i]]
                if(start < hash[s[i]])
                    start = hash[s[i]];
                len = max(i - start, len);
                hash[s[i]] = i;
            }   

            return len;
        }   
};