Lintcode: Previous Permuation

Given a list of integers, which denote a permutation.

Find the previous permutation in ascending order.

Note
The list may contains duplicate integers.

Example
For [1,3,2,3], the previous permutation is [1,2,3,3]

For [1,2,3,4], the previous permutation is [4,3,2,1]

跟Next Permutation很像，只不过条件改成

for (int i=nums.lenth-2; i>=0; i--)

　　if (nums[i] > nums[i+1]) break;

for (int j=i; j<num.length-1; j++)

　　if (nums[j+1]>=nums[i]) break;

 1 public class Solution {
 2     /**
 3      * @param nums: A list of integers
 4      * @return: A list of integers that‘s previous permuation
 5      */
 6     public ArrayList<Integer> previousPermuation(ArrayList<Integer> nums) {
 7         // write your code
 8         if (nums==null || nums.size()==0) return nums;
 9         int i = nums.size()-2;
10         for (; i>=0; i--) {
11             if (nums.get(i) > nums.get(i+1)) break;
12         }
13         if (i >= 0) {
14             int j=i;
15             for (; j<=nums.size()-2; j++) {
16                 if (nums.get(j+1) >= nums.get(i)) break;
17             }
18             int temp = nums.get(j);
19             nums.set(j, nums.get(i));
20             nums.set(i, temp);
21         }
22         reverse(nums, i+1);
23         return nums;
24     }
25
26     public void reverse(ArrayList<Integer> nums, int k) {
27         int l = k, r = nums.size()-1;
28         while (l < r) {
29             int temp = nums.get(l);
30             nums.set(l, nums.get(r));
31             nums.set(r, temp);
32             l++;
33             r--;
34         }
35     }
36 }