LeetCode 76. Minimum Window Substring（最小窗口子串）

原题网址：https://leetcode.com/problems/minimum-window-substring/

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.

方法一：通过直方图记录。

public class Solution {
    private boolean cover(int[] f1, int[] f2) {
        for(int i=0; i

方法二：引入计数器，避免比较直方图。

public class Solution {
    /*
    这道题最初的性能瓶颈在于每次检查histogram，
    网上搜索了一些方法，原来可以用一个matched计数器，避免检查histogram！
    http://www.cnblogs.com/TenosDoIt/p/3461301.html
    */
    private boolean cover(int[] f1, int[] f2) {
        for(int i=0; i= tf[sa[from-1]]);
                matched --;
            }
        }
        if (minLength < 0) return "";
        return s.substring(minFrom, minFrom + minLength);
    }
}

另一种实现：

public class Solution {
    public String minWindow(String s, String t) {
        if (t.length() == 0) return "";
        char[] ta = t.toCharArray();
        int[] h = new int[256];
        for(char ch : ta) {
            h[ch]++;
        }
        
        char[] sa = s.toCharArray();
        int[] f = new int[256];
        int from = 0;
        int match = 0;
        int minFrom = 0, minSize = -1;
        for(int i = 0; i < sa.length; i++) {
            if (f[sa[i]]++ < h[sa[i]]) {
                match++;
            }
            if (match == ta.length && (minSize == -1 || i - from + 1 < minSize)) {
                minFrom = from;
                minSize = i - from + 1;
            }
            while (match == ta.length || (from <= i && f[sa[from]] > h[sa[from]])) {
                if (f[sa[from]] == h[sa[from]]) match--;
                f[sa[from++]]--;
            }
        }
        if (minSize == -1) return "";
        return s.substring(minFrom, minFrom + minSize);
    }
}