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

76. 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

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
}