Given an arrayAof positive integers (not necessarily distinct), return the lexicographically largest permutation that is smaller thanA, that can bemade with one swap(Aswapexchanges the positions of two numbersA[i]andA[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

class Solution {
    public int[] prevPermOpt1(int[] A) {
        int n = A.length;
        int i = n - 1;
        while(i >= 1) {
            if(A[i - 1] <= A[i]) i--;
            else break;
        }
        if(i == 0) return A;
        i--;
        //System.out.print(i);
        int sec = A[i + 1], j = i + 2;
        
 
int secind = i + 1;
        while(j < n) {
            if(A[j] > sec && A[j] < A[i]) {
                sec = A[j];
                secind = j;
            }
            j++;
        }
        int tmp = A[i];
        A[i] = A[secind];
        A[secind] = tmp;
        return A;
    }
}

如果是升序就無解。1.從右往左找第一個破壞降序的，比如4321，2》1所以破壞了降序，2就是我們要換的數字，

2.從2右邊的數字中找出第一個比他小的，如果相同取左邊，交換即可。