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 <= A.length <= 100001 <= 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:
1 class Solution { 2 public int[] prevPermOpt1(int[] A) { 3 if(A == null || A.length == 0){ 4 return A; 5 } 6 7 int n = A.length; 8 int i = n - 1; 9 while(i >= 1 && A[i] >= A[i - 1]){ 10 i--; 11 } 12 13 if(i == 0){ 14 return A; 15 } 16 17 int first = i - 1; 18 int second = i; 19 for(int j = i + 1; j < n; j++){ 20 if(A[j] > A[j - 1] && A[j] < A[first]){ 21 second = j; 22 } 23 } 24 25 int temp = A[first]; 26 A[first] = A[second]; 27 A[second] = temp; 28 return A; 29 } 30 }