Permutation Sequence

The set [1,2,3,…,n] contains a total of n! unique permutations.

By listing and labeling all of the permutations in order,
We get the following sequence (ie, for n = 3):

1. "123"
2. "132"
3. "213"
4. "231"
5. "312"
6. "321"

Given n and k, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

这道题直接在上一题的基础上修改了一下。上一题找出next permutation。只要循环找出k次就可以了，next permutation时间复杂度有点高，还需要优化一下

public class Solution {
    
    public String getPermutation(int n, int k) {
        int num[] = new int[n];
        for(int i = 1; i <= num.length; i++){
            num[i - 1] = i;
        }//初始化数组
        for(int i = 1; i < k; i++){
            nextPermutation(num);
        }//for
        String result = "";
        for(int i = 0; i < num.length; i++){
            result += String.valueOf(num[i]);
        }
        
        return result;
    }
    
    /**
     * 获取下一个排列
     * @param num
     */
    private void nextPermutation(int[] num) {
        if(num.length == 0 || num.length == 1)
            return;
        boolean found = false;
        int index = 0;
        int comparaValue = 0;
        for(int i = num.length - 1; i > 0; i--){        //找出出现降序的位置
            if(num[i] > num[i - 1]){                    //这里不能交换
                index = i;
                comparaValue = i - 1;
                break;
            }//if
        }//for

            int min = index;
            for(int i = min; i < num.length; i++){
                if(num[i] > num[comparaValue] && num[i] < num[min]){
                    min = i;
                }//if                
            }//for
            int temp = num[min];
            num[min] = num[comparaValue];
            num[comparaValue] = temp;
            
            //对index后面的进行升序排列
            for(int i = index; i < num.length; i++){
                min = i;
                for(int j = i + 1; j < num.length; j++){
                    if(num[min] > num[j])
                        min = j;
                }//for
                if(min != i){
                    temp = num[min];
                    num[min] = num[i];
                    num[i] = temp;
                }
            }
        
    }
}