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.
https://oj.leetcode.com/problems/minimum-window-substring/
思路：窗口移动法。 详见这里
public class Solution { public String minWindow(String S, String T) { if (S == null || T == null) return ""; int[] needToFind = new int[256]; int[] found = new int[256]; for (int i = 0; i < T.length(); i++) needToFind[T.charAt(i)]++; String result = ""; int begin = 0, end = 0; int minWin = Integer.MAX_VALUE; int count = 0; for (end = 0; end < S.length(); end++) { if (needToFind[S.charAt(end)] == 0) continue; char ch = S.charAt(end); found[ch]++; if (found[ch] <= needToFind[ch]) count++; if (count == T.length()) { //move the begin pointer while the constrain meets while (begin < S.length() && (found[S.charAt(begin)] > needToFind[S.charAt(begin)] || needToFind[S.charAt(begin)] == 0)) { if (found[S.charAt(begin)] > 0) found[S.charAt(begin)]--; begin++; } //update the min according to the current window if (end - begin + 1 < minWin) { minWin = end - begin + 1; result = S.substring(begin, end + 1); } } } return result; } public static void main(String[] args) { System.out.println(new Solution().minWindow("ADOBECODEBANC", "ABC")); } }