leetcode:Minimum Window Substring 细致分析

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.

题目分析：得到字符串S中包含所有T的字母的最小子串，看到此题就想起了滑动窗口左右index处理，以及MAP处理保证所有的字符出现一次。

基本思路如下：

1. 采用MAP把T中的所有字符记录下来，这主要是为了处理重复字符

2. 采用begin,end两个变量记录字符串的开始和结束

3. 当map不够时，end前进一步,每发现一个，对应map值减一

4. 当map足够时，begin前进，每去掉一个，对应map值加一

具体代码如下：

public static String minWindow(String s, String t) {String res = new String();if (t == null || s == null)return res;if (t.length() <= 0 || s.length() <= 0)return res;Map<Character, Integer> map = new HashMap<Character, Integer>();// 获取对应字符串mapfor (int i = 0; i < t.length(); i++) {char ch = t.charAt(i);if (map.containsKey(ch)) {map.put(ch, map.get(ch) + 1);} else {map.put(ch, 1);}}int begin = 0;int end = 0;int minlen = Integer.MAX_VALUE;int minbegin = -1;int minend = -1;//map不够,index++while (end < s.length()) {char ch2 = s.charAt(end);if (map.containsKey(ch2)) {map.put(ch2, map.get(ch2) - 1);if (judge(map)) {int winlen = end - begin + 1;if (minlen > winlen) {minlen = winlen;minbegin = begin;minend = end;}if(minlen==t.length())  break;//map足够，begin++while (begin <= end && judge(map)) {if (judge(map)) {int blen = end - begin + 1;if (minlen > blen) {minlen = blen;minbegin = begin;minend = end;}}char ch3 = s.charAt(begin);if (map.containsKey(ch3)) {map.put(ch3, map.get(ch3) + 1);}begin++;}}}end++;}if (minend == -1)return res;elsereturn s.substring(minbegin, minend + 1);}public static boolean judge(Map<Character, Integer> map) {for (Integer value : map.values()) {if (value > 0)return false;}return true;}@Testpublic void case1() {String s = "ADOBECODEBANC";String t = "ABC";String actual = minWindow(s, t);String result = "BANC";Assert.assertEquals(result, actual);}@Testpublic void case2() {String s = "BDAB";String t = "AB";String actual = minWindow(s, t);String result = "AB";Assert.assertEquals(result, actual);}