Minimum Window Substring(Hard)最小窗口子串

LeetCode Source

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 empty string "".

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

解题思路：双指针+双HashMap

Java代码

public class Solution {
    public String minWindow(String S, String T) {
        HashMap<Character, Integer> dict = new HashMap<>();
        for (int i = 0; i < T.length(); i++) {
            char c = T.charAt(i);
            if (!dict.containsKey(c))
                dict.put(c, 1);
            else
                dict.put(c, dict.get(c) + 1);
        }
        HashMap<Character, Integer> found = new HashMap<>();
        int foundCounter = 0;
        int start = 0;
        int end = 0;
        int min = Integer.MAX_VALUE;
        String minWindow = "";
        while (end < S.length()) {
            char c = S.charAt(end);
            if (dict.containsKey(c)) {
                if (found.containsKey(c)) {
                    if (found.get(c) < dict.get(c))
                        foundCounter++;
                    found.put(c, found.get(c) + 1);
                } else {
                    found.put(c, 1);
                    foundCounter++;
                }
            }
            if (foundCounter == T.length()) {
        //When foundCounter equals to T.length(), in other words, found all characters.
                char sc = S.charAt(start);
                while (!found.containsKey(sc) || found.get(sc) > dict.get(sc)) {
                    if (found.containsKey(sc) && found.get(sc) > dict.get(sc))
                        found.put(sc, found.get(sc) - 1);
                    start++;
                    sc = S.charAt(start);
                }
                if (end - start + 1 < min) {
                    minWindow = S.substring(start, end + 1);
                    min = end - start + 1;
                }
            }
            end++;
        }
        return minWindow;
    }
}

References

1 没有重复字符的最长子串

2 Minimum Window Substring from Life in Code