76. Minimum Window Substring

难度: Hard

题目描述

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?

解题思路

代码实现

解决方案

java

class Solution {
    public String minWindow(String s, String t) {
        if (s == null || s.isEmpty() || s.length() < t.length()) {
            return "";
        }

        Map<Character, Integer> need = new HashMap<>();
        for (char c : t.toCharArray()) {
            need.put(c, need.getOrDefault(c, 0) + 1);
        }

        int left = 0;
        int right = 0;
        int validCount = 0;
        int start = 0;
        int len = Integer.MAX_VALUE;

        Map<Character, Integer> window = new HashMap<>();

        while (right < s.length()) {
            char c = s.charAt(right);
            right++;

            if (need.containsKey(c)) {
                window.put(c, window.getOrDefault(c, 0) + 1);
                if (window.get(c).equals(need.get(c))) {
                    validCount++;
                }
            }

            while (validCount == need.size()) {
                if (right - left < len) {
                    start = left;
                    len = right - left;
                }

                char d = s.charAt(left);
                left++;

                if (need.containsKey(d)) {
                    if (window.get(d).equals(need.get(d))) {
                        validCount--;
                    }
                    window.put(d, window.get(d) - 1);
                }
            }
        }

        return len == Integer.MAX_VALUE ? "" : s.substring(start, start + len);
    }
}