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(target int, nums []int) int {
	left, sum, res := 0, 0, len(nums)+1
	for right, v := range nums {
		sum += v
		for sum >= target {
			res = min(res, right-left+1)
			sum -= nums[left]
			left++
		}
	}
	if res == len(nums)+1 {
		return 0
	}
	return res
}

func min(a int, b int) int {
	if a > b {
		return b
	}
	return a
}