[Swift]LeetCode1053.交换一次的先前排列 | Previous Permutation With One Swap

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Given an array A of positive integers (not necessarily distinct), return the lexicographically largest permutation that is smaller than A, that can be made with one swap (A swap exchanges the positions of two numbers A[i] and A[j]).  If it cannot be done, then return the same array.

Example 1:

Input: [3,2,1]
Output: [3,1,2]
Explanation: 
Swapping 2 and 1.

Example 2:

Input: [1,1,5]
Output: [1,1,5]
Explanation: 
This is already the smallest permutation.

Example 3:

Input: [1,9,4,6,7]
Output: [1,7,4,6,9]
Explanation: 
Swapping 9 and 7.

Example 4:

Input: [3,1,1,3]
Output: [1,1,3,3]

Note:

1. 1 <= A.length <= 10000
2. 1 <= A[i] <= 10000

---

 给你一个正整数的数组 A（其中的元素不一定完全不同），请你返回可在 一次交换（交换两数字 A[i] 和 A[j] 的位置）后得到的、按字典序排列小于 A 的最大可能排列。

如果无法这么操作，就请返回原数组。

示例 1：

输入：[3,2,1]
输出：[3,1,2]
解释：
交换 2 和 1

示例 2：

输入：[1,1,5]
输出：[1,1,5]
解释： 
这已经是最小排列

示例 3：

输入：[1,9,4,6,7]
输出：[1,7,4,6,9]
解释：
交换 9 和 7

示例 4：

输入：[3,1,1,3]
输出：[1,1,3,3]

提示：

1. 1 <= A.length <= 10000
2. 1 <= A[i] <= 10000

---

Runtime: 264 ms

Memory Usage: 21.7 MB

class Solution {
    func prevPermOpt1(_ A: [Int]) -> [Int] {
        var A = A
        let n:Int = A.count
        for i in stride(from:n - 1,to:0,by:-1)
        {
            if A[i-1] <= A[i] {continue}
            let id:Int = i - 1
            for j in stride(from:n - 1,to:id,by:-1)
            {
                if A[id] <= A[j] {continue}
                A.swapAt(id,j)
                return A
            }
        }
        return A
    }
}

---

268ms

class Solution {
    func prevPermOpt1(_ A: [Int]) -> [Int] {
        var A = A, i = A.count - 2
        while i >= 0 && A[i] <= A[i+1] {
            i -= 1
        }
        
        guard i >= 0 else { return A }
        
        var lo = i + 1, hi = A.count
        while lo < hi {
            let mid = lo + (hi - lo) / 2
            if A[i] > A[mid] {
                lo = mid + 1
            } else {
                hi = mid
            }
        }
        
        var target = lo - 1
        while target > i && A[target] == A[target-1] {
            target -= 1
        }
        A.swapAt(i, target)
        return A
    }
}

---

276ms

class Solution {
    func prevPermOpt1(_ A: [Int]) -> [Int] {
        var A = A, i = A.count - 2
        while i >= 0 && A[i] <= A[i+1] {
            i -= 1
        }
        
        guard i >= 0 else { return A }
        
        var lo = i + 1, hi = A.count
        while lo < hi {
            let mid = lo + (hi - lo) / 2
            if A[i] > A[mid] {
                lo = mid + 1
            } else {
                hi = mid
            }
        }
        
        A.swapAt(i, lo - 1)
        return A
    }
}