Lintcode: Previous Permuation

Given a list of integers, which denote a permutation.Find the previous permutation in ascending order.NoteThe list may contains duplicate integers.ExampleFor [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 }