LeetCode 1053. Previous Permutation With One Swap

原题链接在这里：https://leetcode.com/problems/previous-permutation-with-one-swap/

题目：

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: [1,3,1,3]
Explanation: Swapping 1 and 3.

Note:

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

题解：

From right to left, find the first element A[i - 1] > A[i]. Mark first position = i - 1.

If we can't find it, then there is no previouis perrmutation.

Then mark second position = i, and for j > i, find the last position that A[j] > A[j - 1] && A[j] < A[first]. If we find any, keep updating second position.

Swap first and second positions.

Time Complexity: O(n). n = A.length.

Space: O(1).

AC Java:

class Solution {
    public int[] prevPermOpt1(int[] A) {
        if(A == null || A.length == 0){
            return A;
        }
        
        int n = A.length;
        int i = n - 1;
        while(i >= 1 && A[i] >= A[i - 1]){
            i--;
        }
        
        if(i == 0){
            return A;
        }
        
        int first = i - 1;
        int second = i;
        for(int j = i + 1; j < n; j++){
            if(A[j] > A[j - 1] && A[j] < A[first]){
                second = j;
            }
        }
            
        int temp = A[first];
        A[first] = A[second];
        A[second] = temp;
        return A;
    }
}

类似Next Permutation.