面试经典150题——Day33

一、题目

76. Minimum Window Substring

Given two strings s and t of lengths m and n respectively, return the minimum window
substring
of s such that every character in t (including duplicates) is included in the window. If there is no such substring, return the empty string “”.

The testcases will be generated such that the answer is unique.

Example 1:

Input: s = “ADOBECODEBANC”, t = “ABC”
Output: “BANC”
Explanation: The minimum window substring “BANC” includes ‘A’, ‘B’, and ‘C’ from string t.
Example 2:

Input: s = “a”, t = “a”
Output: “a”
Explanation: The entire string s is the minimum window.
Example 3:

Input: s = “a”, t = “aa”
Output: “”
Explanation: Both 'a’s from t must be included in the window.
Since the largest window of s only has one ‘a’, return empty string.

Constraints:

m == s.length
n == t.length
1 <= m, n <= 105
s and t consist of uppercase and lowercase English letters.

Follow up: Could you find an algorithm that runs in O(m + n) time?

题目来源： leetcode

二、题解

滑动窗口，两个哈希表

class Solution {
public:
    string minWindow(string s, string t) {
        string res = s;
        int n1 = s.length(),n2 = t.length();
        if(n1 < n2) return "";
        unordered_map<char,int> tmap;
        unordered_map<char,int> window;
        for(int i = 0;i < n2;i++) tmap[t[i]]++;
        int i = 0,j = 0;
        int begin = 0;
        int resLen = s.length();
        int cnt = 0;
        bool exist = false;
        while(j < n1){
            //如果未满，则扩大j
            if(tmap[s[j]] == 0){
                j++;
                continue;
            }
            if(window[s[j]] < tmap[s[j]]) cnt++;
            window[s[j]]++;
            j++;
            cout << 111 << endl;
            while(cnt == n2){
                if(j - i <= resLen){
                    begin = i;
                    resLen = j - i;
                    exist = true;
                }
                //缩小i
                if(tmap[s[i]] == 0){
                    i++;
                    continue;
                }
                if(window[s[i]] == tmap[s[i]]){
                    cnt--;
                }
                window[s[i]]--;
                i++;
            }
        }
        res = s.substr(begin,resLen);
        return exist ? res : "";
    }
};
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