mirror of
https://github.com/halfrost/LeetCode-Go.git
synced 2025-08-03 02:47:26 +08:00
77 lines
2.0 KiB
Markdown
77 lines
2.0 KiB
Markdown
# [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
|
||
}
|
||
|
||
```
|
||
|