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?
class Solution {
public String minWindow(String s, String t) {
// 特殊情况直接返回
if(s == null || t == null || s.length() == 0 || t.length() == 0 || s.length() < t.length()) {
return "";
}
// 构造模型
int minLeft = 0;
int minRight = 0;
int min = s.length();
boolean flag = false;
Map<Character, Integer> map = new HashMap<>();
int count = t.length(); // 记录t中需要匹配的字符数
for(char c : t.toCharArray()){
map.put(c, map.getOrDefault(c, 0) + 1);
}
// 滑动窗口解决问题
int i = 0;
int j = 0;
while(j < s.length()){
char c = s.charAt(j);
if(map.containsKey(c)){
map.put(c, map.get(c) - 1);
// 如果仍然有未匹配的字符，count-1
if(map.get(c) >= 0) count--;
}
// 如果发现了一个子串
while(count == 0 && i <= j){
// 更新全局最小值
flag = true;
int curLen = j + 1 - i;
if(curLen <= min){
minLeft = i;
minRight = j;
min = curLen;
}
// 收缩左侧滑动窗口
char leftC = s.charAt(i);
if(map.containsKey(leftC)){
map.put(leftC, map.get(leftC) + 1);
if(map.get(leftC) >= 1) count++;
}
i++;
}
j++;
}
return flag == true ? s.substring(minLeft, minRight + 1): "";
}
}