16.LeetCode刷题For Swift·209. Mini

1、原题

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:

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).

2、思路

3、代码

class Solution {
    func minSubArrayLen(_ s: Int, _ nums: [Int]) -> Int {
        // 结果，初始化为一个超大值，不能为空
        var result = INTPTR_MAX
        // 左指针
        var i = 0
        // 求和
        var sum = 0
        // 最小长度
        var subLength = 0
        for j in 0..<nums.count {
            // 求和，判断和是否大于等于s
            sum += nums[j]
            while sum >= s {
                // 小数组的长度
                subLength = j - i + 1
                result = result < subLength ? result : subLength
                // 然后还得减去小数组的第一个元素
                sum -= nums[i]
                i += 1
            }
        }
        return result == INTPTR_MAX ? 0 : result
    }
}