[Swift]LeetCode1053.交換一次的先前排列 | 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

---

 給你一個正整數的數組 A（其中的元素不必定徹底不一樣），請你返回可在 一次交換（交換兩數字 A[i] 和 A[j] 的位置）後獲得的、按字典序排列小於 A 的最大可能排列。

若是沒法這麼操做，就請返回原數組。

示例 1：

輸入：[3,2,1]
輸出：[3,1,2]
解釋：
交換 2 和 1

示例 2：

輸入：[1,1,5]
輸出：[1,1,5]
解釋： 
這已是最小排列

示例 3：

輸入：[1,9,4,6,7]
輸出：[1,7,4,6,9]
解釋：
交換 9 和 7

示例 4：

輸入：[3,1,1,3]
輸出：[1,1,3,3]

提示：

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

---

Runtime: 264 ms

Memory Usage: 21.7 MB

 1 class Solution {
 2     func prevPermOpt1(_ A: [Int]) -> [Int] {
 3         var A = A
 4         let n:Int = A.count
 5         for i in stride(from:n - 1,to:0,by:-1)
 6         {
 7             if A[i-1] <= A[i] {continue}
 8             let id:Int = i - 1
 9             for j in stride(from:n - 1,to:id,by:-1)
10             {
11                 if A[id] <= A[j] {continue}
12                 A.swapAt(id,j)
13                 return A
14             }
15         }
16         return A
17     }
18 }

---

268ms

 

 1 class Solution {
 2     func prevPermOpt1(_ A: [Int]) -> [Int] {
 3         var A = A, i = A.count - 2
 4         while i >= 0 && A[i] <= A[i+1] {
 5             i -= 1
 6         }
 7         
 8         guard i >= 0 else { return A }
 9         
10         var lo = i + 1, hi = A.count
11         while lo < hi {
12             let mid = lo + (hi - lo) / 2
13             if A[i] > A[mid] {
14                 lo = mid + 1
15             } else {
16                 hi = mid
17             }
18         }
19         
20         var target = lo - 1
21         while target > i && A[target] == A[target-1] {
22             target -= 1
23         }
24         A.swapAt(i, target)
25         return A
26     }
27 }

---

276ms

 1 class Solution {
 2     func prevPermOpt1(_ A: [Int]) -> [Int] {
 3         var A = A, i = A.count - 2
 4         while i >= 0 && A[i] <= A[i+1] {
 5             i -= 1
 6         }
 7         
 8         guard i >= 0 else { return A }
 9         
10         var lo = i + 1, hi = A.count
11         while lo < hi {
12             let mid = lo + (hi - lo) / 2
13             if A[i] > A[mid] {
14                 lo = mid + 1
15             } else {
16                 hi = mid
17             }
18         }
19         
20         A.swapAt(i, lo - 1)
21         return A
22     }
23 }