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61 lines
1.5 KiB
Markdown
61 lines
1.5 KiB
Markdown
# [209. Minimum Size Subarray Sum](https://leetcode.com/problems/minimum-size-subarray-sum/)
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## 题目
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Given an array of n positive integers and a positive integer s, find the minimal length of a contiguous subarray of which the sum ≥ s. If there isn't one, return 0 instead.
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**Example 1**:
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```
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Input: s = 7, nums = [2,3,1,2,4,3]
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Output: 2
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Explanation: the subarray [4,3] has the minimal length under the problem constraint.
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```
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**Follow up**:
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If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).
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## 题目大意
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给定一个整型数组和一个数字 s，找到数组中最短的一个连续子数组，使得连续子数组的数字之和 sum>=s，返回最短的连续子数组的返回值。
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## 解题思路
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这一题的解题思路是用滑动窗口。在滑动窗口 [i,j]之间不断往后移动，如果总和小于 s，就扩大右边界 j，不断加入右边的值，直到 sum > s，之和再缩小 i 的左边界，不断缩小直到 sum < s，这时候右边界又可以往右移动。以此类推。
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## 代码
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```go
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package leetcode
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func minSubArrayLen(s int, nums []int) int {
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n := len(nums)
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if n == 0 {
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return 0
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}
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left, right, res, sum := 0, -1, n+1, 0
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for left < n {
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if (right+1) < n && sum < s {
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right++
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sum += nums[right]
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} else {
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sum -= nums[left]
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left++
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}
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if sum >= s {
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res = min(res, right-left+1)
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}
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}
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if res == n+1 {
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return 0
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}
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return res
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}
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``` |