[Swift]LeetCode862. 和至少为 K 的最短子数组 | Shortest Subarray with Sum at Least K

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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 的最短的非空连续子数组的长度，该子数组的和至少为 K 。

如果没有和至少为 K 的非空子数组，返回 -1 。

 示例 1：

输入：A = [1], K = 1
输出：1

示例 2：

输入：A = [1,2], K = 4
输出：-1

示例 3：

输入：A = [2,-1,2], K = 3
输出：3 

提示：

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

---

956ms

class Solution {
    func shortestSubarray(_ A: [Int], _ K: Int) -> Int {
        var prefixSums = [Int]()
        prefixSums.append(0)
        var sum = 0
        for (i, num) in A.enumerated() {
            sum += num
            prefixSums.append(sum)
        }
        var res = Int.max
        var deque = Deque<(Int, Int)>()
        for i in 0..<prefixSums.count {
            while !deque.isEmpty && deque.back.0 >= prefixSums[i] {
                deque.popBack()
            }
            deque.push( (prefixSums[i], i) )
            while !deque.isEmpty && prefixSums[i] - deque.front.0 >= K {
                res = min(res, i - deque.front.1)
                deque.popHead()
            }
        }
        return res == Int.max ? -1 : res
    }
}

struct Deque<T> {
    var head = 0
    var tail = 0
    var arr = [T]()
    
    mutating func push(_ val: T) {
        arr.append(val)
        tail += 1
    }
    
    @discardableResult mutating func popBack() -> T {
        let val = arr.removeLast()
        tail -= 1
        return val
    }
    
    @discardableResult mutating func popHead() -> T {
        let val = arr[head]
        head += 1
        return val
    }
    
    var front: T {
        return arr[head]
    }
    
    var back: T {
        return arr[tail - 1]
    }
    
    var isEmpty: Bool {
        return head == tail
    }
}

---

Runtime: 1240 ms

Memory Usage: 20.6 MB

class Solution {
    func shortestSubarray(_ A: [Int], _ K: Int) -> Int {
        var N:Int = A.count
        var res:Int = N + 1
        var B:[Int] = [Int](repeating:0,count:N + 1)
        for i in 0..<N
        {
            B[i + 1] = B[i] + A[i]
        }
        var d:[Int] = [Int]()
        for i in 0..<(N + 1)
        {
            while (d.count > 0 && B[i] - B[d.first!] >= K)
            {
                res = min(res, i - d.first!)
                 d.removeFirst()
            }
            while (d.count > 0 && B[i] <= B[d.last!])
            {
                d.popLast()
            }
            d.append(i)
        }
        return res <= N ? res : -1
    }
}