209. Minimum Size Subarray Sum

Medium

Given an array of positive integers nums and a positive integer target, return the minimal length of a

subarray whose sum is greater than or equal to target. If there is no such subarray, return 0 instead.

Example 1:

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

Example 2:

Input: target = 4, nums = [1,4,4]
Output: 1

Example 3:

Input: target = 11, nums = [1,1,1,1,1,1,1,1]
Output: 0

Constraints:

• 1 <= target <= 10^9

• 1 <= nums.length <= 10^5

• 1 <= nums[i] <= 10^4

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)).

解題

O(n) 的 solution, Sliding window

func minSubArrayLen(target int, nums []int) int {
    res := 1000000
    sum := 0
    left := 0

    for i:=0; i<len(nums); i++ {
        sum += nums[i]


        if sum >= target { // 加上去 window 右邊的值大於目標
            
            for { //將 window 左邊一直右移(縮小 window)  直到不能再移(再移sum會小於目標)
                if left >= 0 && sum-nums[left] >= target {
                    sum -= nums[left]
                    left++
                } else {
                    break
                }
            }

            res = min(res, i-left+1)
            if res == 1 { return res } //除非無解，不然不會有比 1 更小的答案
        }
    }

    if res==1000000 { //無解
        return 0
    }

    return res
}

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

    return b
}