[LeetCode][Java] Minimum Window Substring

题目：

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.

题意：

给定一个字符串S 和一个字符串T,在字符串S中找出最小的窗口，这个窗口可以包含T中的所有的字符，，时间复杂度要求为O(n).

举个例子：

S="ADOBECODEBANC"T="ABC"

最下的窗口为"BANC".

算法分析：

// 双指针思想，尾指针不断往后扫，当扫到有一个窗口包含了所有T的字符，然后再收缩头指针，直到不能再收缩为止。// 最后记录所有可能的情况中窗口最小的

AC代码：

public class Solution {public String minWindow(String S, String T){HashMap<Character, Integer> hasFound = new HashMap<Character, Integer>();HashMap<Character, Integer> needToFind = new HashMap<Character, Integer>();for (int i = 0; i < T.length(); i++){hasFound.put(T.charAt(i), 0);if (needToFind.containsKey(T.charAt(i))){needToFind.put(T.charAt(i), needToFind.get(T.charAt(i)) + 1);}else{needToFind.put(T.charAt(i), 1);}}int begin = 0;int minWindowSize = S.length();String retString = "";int count = 0;for (int end = 0; end < S.length(); end++){Character end_c = S.charAt(end);if (needToFind.containsKey(end_c)){hasFound.put(end_c, hasFound.get(end_c) + 1);if (hasFound.get(end_c) <= needToFind.get(end_c)){count++;}if (count == T.length()){while ((!needToFind.containsKey(S.charAt(begin)))||(hasFound.get(S.charAt(begin)) > needToFind.get(S.charAt(begin)))){if (needToFind.containsKey(S.charAt(begin))){hasFound.put(S.charAt(begin),hasFound.get(S.charAt(begin)) – 1);}begin++;}if ((end – begin + 1) <= minWindowSize){minWindowSize = end – begin + 1;retString = S.substring(begin, end + 1);}}}}return retString;}}

人，都有不能称心如意的时候，都有愿望落空的窘迫，

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