19.2.15 [LeetCode 76] Minimum Window Substring

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.

题意

给出两个字符串s和t，要求s中的一个最短子串使得这个子串能覆盖t中所有的字符，如果不存在就输出空字符串

题面其实挺坑的，因为覆盖t中所有的字符意思是，如果t中出现重复字符则输出子串中相应字符的出现次数要大于等于t中的。

这种文字游戏就很没意思了，我一开始按t中字符只要s中出现就算是覆盖的，很坑

题解

class Solution {
public:
    string minWindow(string s, string t) {
        int p1 = 0, p2 = -1, l1 = s.length(), l2 = t.length(), kind = 0;
        vector<int>mark(256, 0);
        vector<int>count(256, 0);
        for (int i = 0; i < l2; i++) {
            mark[t[i]]++;
            kind++;
        }
        int S = -1, E = -1, countkind = 0, minl = INT_MAX;
        while (p2 < l1 - 1) {
            if (mark[s[p2 + 1]]) {
                if (count[s[p2 + 1]] < mark[s[p2 + 1]])countkind++;
                count[s[p2 + 1]]++;
            }
            while (p1 < p2 + 1 && (mark[s[p1]] == 0 || count[s[p1]] > mark[s[p1]])) {
                if (mark[s[p1]])count[s[p1]]--;
                p1++;
            }
            if (kind == countkind && (p2 - p1 + 1) < minl) {
                minl = p2 - p1 + 2;
                S = p1, E = p2;
            }
            p2++;
        }
        if (S == -1)return "";
        return s.substr(S, minl);
    }
};

一道略微复杂的滑动窗口题，想明白了写不难

这道题蛮神奇的，第一次memory usage beat了100%，不知道这到底咋算的