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--;
        }
    }
}