[Swift]LeetCode801. 使序列递增的最小交换次数 | Minimum Swaps To Make Sequences Increasing

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We have two integer sequences A and B of the same non-zero length.

We are allowed to swap elements A[i] and B[i].  Note that both elements are in the same index position in their respective sequences.

At the end of some number of swaps, A and B are both strictly increasing.  (A sequence is strictly increasing if and only if A[0] < A[1] < A[2] < ... < A[A.length - 1].)

Given A and B, return the minimum number of swaps to make both sequences strictly increasing.  It is guaranteed that the given input always makes it possible.

Example:
Input: A = [1,3,5,4], B = [1,2,3,7]
Output: 1
Explanation: 
Swap A[3] and B[3].  Then the sequences are:
A = [1, 3, 5, 7] and B = [1, 2, 3, 4]
which are both strictly increasing.

Note:

• A, B are arrays with the same length, and that length will be in the range [1, 1000].
• A[i], B[i] are integer values in the range [0, 2000].

---

我们有两个长度相等且不为空的整型数组 A 和 B 。

我们可以交换 A[i] 和 B[i] 的元素。注意这两个元素在各自的序列中应该处于相同的位置。

在交换过一些元素之后，数组 A 和 B 都应该是严格递增的（数组严格递增的条件仅为A[0] < A[1] < A[2] < ... < A[A.length - 1]）。

给定数组 A 和 B ，请返回使得两个数组均保持严格递增状态的最小交换次数。假设给定的输入总是有效的。

示例:
输入: A = [1,3,5,4], B = [1,2,3,7]
输出: 1
解释: 
交换 A[3] 和 B[3] 后，两个数组如下:
A = [1, 3, 5, 7] ， B = [1, 2, 3, 4]
两个数组均为严格递增的。

注意:

• A, B 两个数组的长度总是相等的，且长度的范围为 [1, 1000]。
• A[i], B[i] 均为 [0, 2000]区间内的整数。

---

Runtime: 64 ms

Memory Usage: 19.2 MB

class Solution {
    func minSwap(_ A: [Int], _ B: [Int]) -> Int {
        var n1:Int = 0
        var s1:Int = 1
        var n:Int = A.count
        for i in 1..<n
        {
            var n2:Int = Int.max
            var s2:Int = Int.max
            if A[i - 1] < A[i] && B[i - 1] < B[i]
            {
                n2 = min(n2, n1)
                s2 = min(s2, s1 + 1)
            }
            if A[i - 1] < B[i] && B[i - 1] < A[i]
            {
                n2 = min(n2, s1)
                s2 = min(s2, n1 + 1)
            }
            n1 = n2
            s1 = s2
        }
        return min(n1, s1)
    }
}

---

76ms

class Solution {
    func minSwap(_ A: [Int], _ B: [Int]) -> Int {                
        var swapRecord = 1, fixRecord = 0
        for i in 1..<A.count {
            if A[i-1] >= B[i] || B[i-1] >= A[i] {
                swapRecord += 1
            }
            else if A[i - 1] >= A[i] || B[i - 1] >= B[i] {
                let temp = swapRecord
                swapRecord = fixRecord + 1
                fixRecord = temp
            }
            else {
                let minv = min(swapRecord, fixRecord)
                swapRecord = minv + 1
                fixRecord = minv
            }
        }
        return min(swapRecord, fixRecord)
    }

    func minSwapDP(_ A: [Int], _ B: [Int], _ i: Int, _ sum: Int) -> Int {

        if i >= A.count {
            return sum
        }

        if i == 0 { return minSwapDP(A, B, i+1, sum) }

        if A[i] > A[i-1] && B[i] > B[i-1] {
            return minSwapDP(A, B, i+1, sum)
        }

        var canSwapPre = (A[i-1] < B[i] && B[i-1] < A[i])
        if canSwapPre {
            for k in 0..<i-1 {
                if A[i-1] <= B[k] || B[i-1] <= A[k]  {
                    canSwapPre = false
                    break
                }
            }
        }
        let canSwapCur = (B[i] > A[i-1] && A[i] > B[i-1])

        var A2 = A, B2 = B
        let tmp2 = A2[i-1]
        A2[i-1] = B2[i-1]
        B2[i-1] = tmp2

        var A1 = A, B1 = B
        let tmp = A1[i]
        A1[i] = B1[i]
        B1[i] = tmp
        if canSwapPre && canSwapCur {
            return min(minSwapDP(A2, B2, i, sum+1), minSwapDP(A1, B1, i+1, sum+1))
        }
        else if canSwapCur {
            return minSwapDP(A1, B1, i+1, sum+1)
        }
        else {
            return minSwapDP(A2, B2, i, sum+1)
        }
    }
}