Return the length of the shortest, non-empty, contiguous subarray of A with sum at least K.

If there is no non-empty subarray with sum at least K, return -1.

Example 1:

Input: A = [1], K = 1
Output: 1

Example 2:

Input: A = [1,2], K = 4
Output: -1

Example 3:

Input: A = [2,-1,2], K = 3
Output: 3
 

Note:

1. 1 <= A.length <= 50000
2. -10 ^ 5 <= A[i] <= 10 ^ 5
3. 1 <= K <= 10 ^ 9

返回 A 的最短的非空連續子數組的長度，該子數組的和至少為 K 。

如果沒有和至少為 K 的非空子數組，返回 -1 。

示例 1：

輸入：A = [1], K = 1
輸出：1

示例 2：

輸入：A = [1,2], K = 4
輸出：-1

示例 3：

輸入：A = [2,-1,2], K = 3
輸出：3 

提示：

1. 1 <= A.length <= 50000
2. -10 ^ 5 <= A[i] <= 10 ^ 5
3. 1 <= K <= 10 ^ 9

 1 class Solution {
 2     func shortestSubarray(_ A: [Int], _ K: Int) -> Int {
 3         var prefixSums = [Int]()
 4         prefixSums.append(0)
 5         var sum = 0
 6         for (i, num) in A.enumerated() {
 7             sum += num
 8             prefixSums.append(sum)
 
 9         }
10         var res = Int.max
11         var deque = Deque<(Int, Int)>()
12         for i in 0..<prefixSums.count {
13             while !deque.isEmpty && deque.back.0 >= prefixSums[i] {
14                 deque.popBack()
15             }
16             deque.push( (prefixSums[i], i) )
17             while !deque.isEmpty && prefixSums[i] - deque.front.0 >= K {
18                 res = min(res, i - deque.front.1)
19                 deque.popHead()
20             }
21         }
22         return res == Int.max ? -1 : res
23     }
24 }
25 
26 struct Deque<T> {
27     var head = 0
28     var tail = 0
29     var arr = [T]()
30     
31     mutating func push(_ val: T) {
32         arr.append(val)
33         tail += 1
34     }
35     
36     @discardableResult mutating func popBack() -> T {
37         let val = arr.removeLast()
38         tail -= 1
39         return val
40     }
41     
42     @discardableResult mutating func popHead() -> T {
43         let val = arr[head]
44         head += 1
45         return val
46     }
47     
48     var front: T {
49         return arr[head]
50     }
51     
52     var back: T {
53         return arr[tail - 1]
54     }
55     
56     var isEmpty: Bool {
57         return head == tail
58     }
59 }

 1 class Solution {
 2     func shortestSubarray(_ A: [Int], _ K: Int) -> Int {
 3         var N:Int = A.count
 4         var res:Int = N + 1
 5         var B:[Int] = [Int](repeating:0,count:N + 1)
 6         for i in 0..<N
 7         {
 8             B[i + 1] = B[i] + A[i]
 9         }
10         var d:[Int] = [Int]()
11         for i in 0..<(N + 1)
12         {
13             while (d.count > 0 && B[i] - B[d.first!] >= K)
14             {
15                 res = min(res, i - d.first!)
16                  d.removeFirst()
17             }
18             while (d.count > 0 && B[i] <= B[d.last!])
19             {
20                 d.popLast()
21             }
22             d.append(i)
23         }
24         return res <= N ? res : -1
25     }
26 }

[Swift]LeetCode862. 和至少為 K 的最短子數組 | Shortest Subarray with Sum at Least K