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;

 public class Solution {
     /**
      * @param nums: A list of integers
      * @return: A list of integers that's previous permuation
      */
     public ArrayList<Integer> previousPermuation(ArrayList<Integer> nums) {
         // write your code
         if (nums==null || nums.size()==0) return nums;
         int i = nums.size()-2;
         for (; i>=0; i--) {
             if (nums.get(i) > nums.get(i+1)) break;
         }
         if (i >= 0) {
             int j=i;
             for (; j<=nums.size()-2; j++) {
                 if (nums.get(j+1) >= nums.get(i)) break;
             }
             int temp = nums.get(j);
             nums.set(j, nums.get(i));
             nums.set(i, temp);
         }
         reverse(nums, i+1);
         return nums;
     }

     public void reverse(ArrayList<Integer> nums, int k) {
         int l = k, r = nums.size()-1;
         while (l < r) {
             int temp = nums.get(l);
             nums.set(l, nums.get(r));
             nums.set(r, temp);
             l++;
             r--;
         }
     }
 }