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 }