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

 1 class Solution {
 2 public:
 3     string minWindow(string s, string t) {
 4         int p1 = 0, p2 = -1, l1 = s.length(), l2 = t.length(), kind = 0;
 5         vector<int>mark(256, 0);
 6         vector<int>count(256, 0);
 7         for (int i = 0; i < l2; i++) {
 8             mark[t[i]]++;
 9             kind++;
10         }
11         int S = -1, E = -1, countkind = 0, minl = INT_MAX;
12         while (p2 < l1 - 1) {
13             if (mark[s[p2 + 1]]) {
14                 if (count[s[p2 + 1]] < mark[s[p2 + 1]])countkind++;
15                 count[s[p2 + 1]]++;
16             }
17             while (p1 < p2 + 1 && (mark[s[p1]] == 0 || count[s[p1]] > mark[s[p1]])) {
18                 if (mark[s[p1]])count[s[p1]]--;
19                 p1++;
20             }
21             if (kind == countkind && (p2 - p1 + 1) < minl) {
22                 minl = p2 - p1 + 2;
23                 S = p1, E = p2;
24             }
25             p2++;
26         }
27         if (S == -1)return "";
28         return s.substr(S, minl);
29     }
30 };
View Code

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

posted @ 2019-02-15 15:16  TobicYAL  阅读(120)  评论(0)  编辑  收藏  举报