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website/content/ChapterFour/0076.Minimum-Window-Substring.md

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+# [76. Minimum Window Substring](https://leetcode.com/problems/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).
+
+**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.
+
+## 题目大意
+
+给定一个源字符串 s，再给一个字符串 T，要求在源字符串中找到一个窗口，这个窗口包含由字符串各种排列组合组成的，窗口中可以包含 T 中没有的字符，如果存在多个，在结果中输出最小的窗口，如果找不到这样的窗口，输出空字符串。
+
+## 解题思路
+
+这一题是滑动窗口的题目，在窗口滑动的过程中不断的包含字符串 T，直到完全包含字符串 T 的字符以后，记下左右窗口的位置和窗口大小。每次都不断更新这个符合条件的窗口和窗口大小的最小值。最后输出结果即可。
+
+
+## 代码
+
+```go
+
+package leetcode
+
+func minWindow(s string, t string) string {
+	if s == "" || t == "" {
+		return ""
+	}
+	var tFreq, sFreq [256]int
+	result, left, right, finalLeft, finalRight, minW, count := "", 0, -1, -1, -1, len(s)+1, 0
+
+	for i := 0; i < len(t); i++ {
+		tFreq[t[i]-'a']++
+	}
+
+	for left < len(s) {
+		if right+1 < len(s) && count < len(t) {
+			sFreq[s[right+1]-'a']++
+			if sFreq[s[right+1]-'a'] <= tFreq[s[right+1]-'a'] {
+				count++
+			}
+			right++
+		} else {
+			if right-left+1 < minW && count == len(t) {
+				minW = right - left + 1
+				finalLeft = left
+				finalRight = right
+			}
+			if sFreq[s[left]-'a'] == tFreq[s[left]-'a'] {
+				count--
+			}
+			sFreq[s[left]-'a']--
+			left++
+		}
+	}
+	if finalLeft != -1 {
+		for i := finalLeft; i < finalRight+1; i++ {
+			result += string(s[i])
+		}
+	}
+	return result
+}
+
+```
+