LeetCode-Go/website/content/ChapterFour/0209.Minimum-Size-Subarray-Sum.md

209. Minimum Size Subarray Sum

题目

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.

Example 1:

Input: s = 7, nums = [2,3,1,2,4,3]
Output: 2
Explanation: the subarray [4,3] has the minimal length under the problem constraint.

Follow up:

If you have figured out the O(n) solution, try coding another solution of which the time complexity is O(n log n).

题目大意

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。

解题思路

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。

代码

package leetcode

func minSubArrayLen(s int, nums []int) int {
	n := len(nums)
	if n == 0 {
		return 0
	}
	left, right, res, sum := 0, -1, n+1, 0
	for left < n {
		if (right+1) < n && sum < s {
			right++
			sum += nums[right]
		} else {
			sum -= nums[left]
			left++
		}
		if sum >= s {
			res = min(res, right-left+1)
		}
	}
	if res == n+1 {
		return 0
	}
	return res
}