題目描述
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.
思路
遍歷t中字符,當符合條件時,左邊界往前走,過程中記錄最小值。
從右邊界進入的字符,哈希表–,num–。左邊界出去的字符,如果當前字符的哈希值==0,num++,哈希表++。
代碼
class Solution {
public:
string minWindow(string s, string t) {
int num = t.length();
unordered_map<char, int> has;
for (char str : t) has[str]++;
int res = INT_MAX;
int left = 0, st = 0;
for (int i=0; i<s.length(); ++i) {
if (has[s[i]]-- > 0) num--;
while(num == 0) {
res = (i - left + 1 < res) ? (i - (st = left) + 1) : res;
if (has[s[left++]]++ == 0) num++;
}
}
return (res == INT_MAX) ? "" : s.substr(st, res);
}
};