【leetcode】Minimum Window Substring (hard) ★

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).

For example,
S =

"ADOBECODEBANC"

T =

"ABC"

Minimum window is

"BANC"

.

Note:
If there is no such window in S that covers all characters in T, return the emtpy string

""

.

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

思路：题目中T可能会有重复的字母 我用了各种分类讨论 结果超时

直接看大神的代码吧

class Solution {
public:
string minWindow(string S, string T) {
int m = S.size(), n = T.size();
if (n <= 0 || m < n) return "";

int require[128] = {0};  //记录每种字母需要多少个 关键点
for (int i = 0; i < n; ++i) require[T[i]]++;

int count = 0;
int minLen = INT_MAX, minIndex = 0;
for (int s = 0, e = 0; e < m; ++e) {
require[S[e]]--; //末尾数字的需求量减1
if (require[S[e]] >= 0) count++;   //如果需求量大于等于0 说明匹配上了新的数字
while (count == n) {  //所有字母都被匹配上了
if (e - s + 1 < minLen) {  //长度变小了 记录下新的长度 和 起始位置
minLen = e - s + 1;
minIndex = s;
}
require[S[s]]++; //起始位置向后移更新需求量
if (require[S[s]] > 0) count--;
s++;
}
}

if (minLen == INT_MAX) return "";
return S.substr(minIndex, minLen);
}
};