[LeetCode]Longest Substring Without Repeating Characters

最新推荐文章于 2025-05-18 01:04:12 发布

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本文介绍了一种用于寻找字符串中最长无重复字符子串的算法实现。通过使用布尔型向量记录字符出现状态，并迭代检查每个字符，一旦发现重复字符即更新最大长度并移动起始位置，从而高效解决问题。

class Solution {
//put last char in the end of the string 
//to avoid extra judge, when go to the end without repeating char
public:
	int lengthOfLongestSubstring(string s) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		if(s.size() == 0) return 0;
		//put last char in the end of the string 
		//to avoid extra judge, when go to the end without repeating char
		s.push_back(s[s.size()-1]);
		vector<bool> cntT(256, false);
		int maxLen = 0;
		int left = 0;
		for (int i = 0; i < s.size(); ++i)
		{
			if(cntT[s[i]] == true)
			{
				if (i-left > maxLen) 
					maxLen = i-left;
				while (s[left] != s[i])
					cntT[s[left++]] = false;
				cntT[s[left++]] = false;
			}
			cntT[s[i]] = true;
		}
		return maxLen;
	}
};

second time

class Solution {
public:
    int lengthOfLongestSubstring(string s) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<bool> charTable(256, false);
        
        int maxLen = 0;
        int startIndex = 0;
        int i;
        for(i = 0; i < s.size(); ++i)
        {
            if(charTable[s[i]] == true)
            {
                maxLen = max(i-startIndex, maxLen);

                while(s[startIndex] != s[i])
                    charTable[s[startIndex++]] = false;
                startIndex++;
            }
            else charTable[s[i]] = true;
        }
           
       return max(maxLen, i-startIndex);
    }
};