面试经典150题 P76 最小覆盖子串

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时间轴

时间轴

2025-11-14

init

滑动窗口

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题目：

P76 最小覆盖子串

https://leetcode.cn/problems/minimum-window-substring/description/?envType=study-plan-v2&envId=top-interview-150

滑动窗口：
window[left, right]，考虑下面问题：

1. 当移动 right 时要更新哪些数据
2. 什么情况下 right 要停止扩大，开始移动 left 缩小窗口
3. 当移动 left 时要更新哪些数据
4. 我们要的结果是在扩大窗口还是缩小窗口时进行更新？

#include <string>
#include <unordered_map>

using std::unordered_map;
using std::string;

class Solution {
    public:
	string minWindow(string s, string t)
	{
		int i = 0, j = 0, n = s.size();
		int start;
		int min_len = n + 1;
		int valid = 0;
		unordered_map<char, int> need;
		unordered_map<char, int> window;

		for (char &ch : t)
			need[ch]++;

		for (j = 0; j < n; j++) {
			if (!need.count(s[j]))
				continue;

			window[s[j]]++;

			if (window[s[j]] == need[s[j]])
				valid++;

			while (valid == need.size()) { // 收缩 left
				if (j - i + 1 < min_len) {
					start = i;
					min_len = j - i + 1;
				}
				if (need.count(s[i])) {
					if (window[s[i]] == need[s[i]])
						valid--;

					window[s[i]]--;
				}
				i++; // 左移窗口
			}
		}
		return min_len == n + 1 ? "" : s.substr(start, min_len);
	}
};

leetcode hot 100 rewrite

#include <string>
#include <unordered_map>

using std::string;
using std::unordered_map;

class Solution {
    public:
        string minWindow(string s, string t)
        {
                int slen = s.size(), tlen = t.size();
                int left = 0, right = 0;
                int valid = 0, total_valid = 0;
                int min_len = slen + 1, res_start = 0;
                unordered_map<char, int> need, window;

                if (tlen > slen)
                        return "";

                for (char ch : t)
                        need[ch]++;

                total_valid = need.size();

                while (right < slen) {
                        window[s[right]]++;
                        if (window[s[right]] == need[s[right]])
                                valid++;
                        right++;

                        while (left < right && valid >= total_valid) { // 收缩

                                if (right - left < min_len) {
                                        min_len = right - left;
                                        res_start = left;
                                }

                                if (window[s[left]] == need[s[left]])
                                        valid--;

                                window[s[left]]--;

                                left++;
                        }
                }
                if(min_len != slen+1)
                        return s.substr(res_start, min_len);
                else
                        return "";
        }
};