159. Longest Substring with At Most Two Distinct Characters

寻找最多包含2个不同字符的子串长度

最新推荐文章于 2024-06-04 16:21:11 发布

原创最新推荐文章于 2024-06-04 16:21:11 发布 · 301 阅读

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本文提供了一种算法，用于在给定字符串中找到最长的子串，该子串最多包含2个不同的字符。通过窗口滑动技巧优化空间复杂度，实现高效计算。

Problem

Given a string, find the length of the longest substring T that contains at most 2 distinct characters.

For example, Given s = “eceba”,

T is "ece" which its length is 3.

Solution

右边每次进来一个字符，并计算以它为最右边界的窗口的最大值。

当加入该字符后，如果数字超过2，那就要考虑去掉当前字符串中的哪个字符之后使得窗口最长，当然是让左边界最小了。

class Solution {
public:
    int lengthOfLongestSubstringTwoDistinct(string s) {
        int rst = 0;
        const int N = s.size(), K = 2;;
        unordered_map<char, int> mp;
        
        for( int left = 0, right = 0; right < s.size(); right++){
            mp[s[right]] = right;
            
            if(mp.size() > K ){
                int rightMost = right;
                char rightMostChar = ' ';
                for( auto it = mp.begin(); it != mp.end(); it++){
                    if(it->second < rightMost){
                        rightMost = it->second;
                        rightMostChar = it->first;
                    }
                }
                mp.erase(rightMostChar);
                left = rightMost + 1;
            }
            
            rst = max(rst, right - left + 1);
        }
        return rst;
    }
};

以上代码适用于K distinct characters.

对于只有两个情况下，可以优化空间复杂度成为常量

class Solution {
public:
    int lengthOfLongestSubstringTwoDistinct(string s) {
        if(s.empty()) return 0;
        
        int rst = 0, j = -1;
        
        int left = 0, right = 0; 
        for( ; right < s.size(); right++){
            if( right == 0 || s[right] == s[right-1]) continue;
            
            if( j >= 0 && s[right] != s[j] && s[right] != s[left]){
                rst = max(rst, right - left);
                left = j + 1;
            }
            j = right - 1;
        }
        return max( right - left, rst);
    }
};