76. Minimum Window Substring

最新推荐文章于 2025-08-18 22:01:01 发布

转载最新推荐文章于 2025-08-18 22:01:01 发布 · 68 阅读

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原文链接：http://www.cnblogs.com/ruruozhenhao/p/9839793.html

文章标签：

#c/c++ #runtime

本文详细解析了如何在O(n)复杂度下找到包含所有目标字符的最小子串，通过C++代码实现并附有具体示例，展示了算法的具体应用。

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Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

Example:

Input: S = "ADOBECODEBANC", T = "ABC"
Output: "BANC"

Note:

If there is no such window in S that covers all characters in T, return the empty string "".
If there is such window, you are guaranteed that there will always be only one unique minimum window in S.

AC code:

class Solution {
public:
    string minWindow(string s, string t) {
        map<char, int> m;
        for (int i = 0; i < t.length(); ++i) 
            m[t[i]]++;
        int total = t.length();
        int from = 0;
        int minn = INT_MAX;
        for (int i = 0, j = 0; i < s.length(); ++i) {
            if (m[s[i]]-- > 0) total--;
            while (total == 0) {
                if (i-j+1 < minn) {
                    minn = i-j+1;
                    from = j;
                }
                if (++m[s[j++]] > 0) total++;    
            }
        }
        return (minn == INT_MAX) ? "" : s.substr(from, minn);
    }
};

Runtime: 20 ms, faster than 45.51% of C++ online submissions for Minimum Window Substring.

c++string.find（）函数用法整理

C++ string substr()

转载于:https://www.cnblogs.com/ruruozhenhao/p/9839793.html