[LintCode] 51. Previous Permutation

public class Solution {
    /*
     * @param nums: A list of integers
     * @return: A list of integers that's previous permuation
     */
       public List<Integer> previousPermuation(List<Integer> nums) {
        int cur = nums.size() - 1;
        for(;cur >=1;cur--){
            //Find the inflection point, if the previous number is greater than the current number, the previous number is the inflection point,
            //Find the largest and smaller number after the inflection point, and exchange it with the inflection point
            if(nums.get(cur - 1) > nums.get(cur)){
                int start = cur;
                int max = nums.get(start);
                int index = start;
                for(;start<nums.size();start++){
                    if(nums.get(start) > max && nums.get(start) < nums.get(cur - 1)){
                        max = nums.get(start);
                        index = start;
                    }
                }
                int temp = nums.get(cur - 1);
                nums.set(cur - 1, nums.get(index));
                nums.set(index, temp);
                
                // Arrange the numbers after the inflection point from largest to smallest
                sortArray(nums,cur);
                break;
            }
        }
        System.out.println(cur);
        if(cur <= 0){
            sortArray(nums,0);
        }
        return nums;
    }
    // Arrange the array from the start position to the smallest
    void sortArray(List<Integer> nums,int start){
        List<Integer> temp = new ArrayList<>();
        for(int i=start;i<nums.size();i++){
            temp.add(nums.get(i));
        }
        Collections.sort(temp);
        int index = 0;
        for(int i=temp.size() - 1;i>=0;i--){
            nums.set(start + index, temp.get(i));
            index ++;
        }
    }
}