Minimum Window Substring LeetCode java

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：双指针，双HashMap

虽然想到了思路，但是写的过程还是特别纠结。。。。。

最后在讨论区找了一段代码：

public String minWindow(String S, String T) {
int[] result = new int[] {-1, S.length()};
int counter = 0;
Map<Character, Integer> expected = new HashMap<>();
Map<Character, Integer> window = new HashMap<>();

for (int i = 0; i < T.length(); i++) {
if (!expected.containsKey(T.charAt(i))) expected.put(T.charAt(i), 0);
expected.put(T.charAt(i), expected.get(T.charAt(i)) + 1);
}
for (int i = 0, j = 0; j < S.length(); j++) {
char cur = S.charAt(j);
if (expected.containsKey(cur)) {
if (!window.containsKey(cur)) window.put(cur, 0);
window.put(cur, window.get(cur) + 1);
if (window.get(cur) <= expected.get(cur)) counter++;
if (counter == T.length()) {
char remove = S.charAt(i);
while (!expected.containsKey(remove) || window.get(remove) > expected.get(remove)){
if (expected.containsKey(remove)) window.put(remove, window.get(remove) - 1);
remove = S.charAt(++i);;
}
if (j - i < result[1] - result[0]) result = new int[]{i, j};
}
}
}
return result[1] - result[0] < S.length() ? S.substring(result[0], result[1] + 1) : "";
}