From 353fecff332693269da020046f1ce857dfc69d76 Mon Sep 17 00:00:00 2001
From: Bugsource <35368347+Bugsource@users.noreply.github.com>
Date: Sun, 18 Aug 2019 19:44:50 +0800
Subject: [PATCH] Create 76. Minimum Window Substring.md

---
 .../76. Minimum Window Substring.md           | 56 +++++++++++++++++++
 1 file changed, 56 insertions(+)
 create mode 100644 sliding window/76. Minimum Window Substring.md

diff --git a/sliding window/76. Minimum Window Substring.md b/sliding window/76. Minimum Window Substring.md
new file mode 100644
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+# 题目描述
+```
+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).
+
+Example:
+
+Input: S = "ADOBECODEBANC", T = "ABC"
+Output: "BANC"
+Note:
+
+If there is no such window in S that covers all characters in T, return the empty string "".
+If there is such window, you are guaranteed that there will always be only one unique minimum window in S.
+```
+# 思路
+1) 关键点在于如何去描述T的窗口，存储出现的字符及个数，最优做法就是用int[256]做map，先存储t里的字符及出现的个数；
+2) 右指针遍历S，将map里对应字符的计数减1，如果map里计数为0，则说明T中不存在，将其减为负数；字符个数大于零，则说明在T中存在，增加找到字符的count；
+3）如果count等于T的长度，说明窗口匹配成功，此时要移动左指针，尽力去缩短匹配的字符串：将map里计数加1，如果是T中没出现过的，最多加到0；如果是出现过的，
+当加到大于0时，说明之前的字符都可以舍弃，此时更新下最短长度；
+
+# 代码
+```
+class Solution {
+    public String minWindow(String s, String t) {
+        int[] window = new int[256];
+        int winSize = t.length();
+        for(int i = 0; i < winSize; i++){
+            window[t.charAt(i)] ++;
+        }
+        int left = 0, len = s.length(), count = 0, minLen = Integer.MAX_VALUE, preMinLeft = 0;
+        for(int right = 0; right < len; right++){
+            char c = s.charAt(right);
+            window[c]--;
+            if(window[c] >= 0){
+                count++;
+            }
+            while(count == winSize){//尝试去移动左指针缩小窗口     
+                char leftChar = s.charAt(left);
+                window[leftChar]++;
+                if(window[leftChar] > 0){//t中没有的但是s中有的，最多加到0；大于0的必然是t中有的                    
+                    count--;
+                    int curLen = right-left+1;
+                    if(curLen < minLen){//循环跳出时才需要更新下最短长度
+                        minLen = curLen;
+                        preMinLeft = left;
+                    }                    
+                } 
+                left++;
+            }
+        }
+        if(minLen == Integer.MAX_VALUE){
+            return "";
+        }
+        return s.substring(preMinLeft, preMinLeft+minLen);      
+    }
+}
+```