leetcode解题方案--060--Permutation Sequence

题目

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):

"123""132""213""231""312""321"

Given n and k, return the kth permutation sequence.

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

分析

我的思路很奇怪
用递归。
- 先判断第k个排序会影响到哪一位
- 得到三个值，分别代表变动的数字个数，前一位增长值和下一次递归的k

class Solution {    public static String getPermutation(int n, int k) {        k = k-1;        int [] cons = new int[11];        cons[0] = 1;        cons[1] = 1;        for (int i = 2; i<=10; i++) {            cons[i] = i*cons[i-1];        }        int[] array = new int[n];        for (int i =0; i<n; i++) {            array[i] = i+1;        }        StringBuffer ret = new StringBuffer("");        getkIndex(ret, array, k, cons);        return ret.toString();    }    public static void getkIndex(StringBuffer xx, int[] array, int k, int[] cons) {        Arrays.sort(array);        int length = array.length;        if (length ==1) {            xx.append(array[0]);            return;        } else if (length == 0) {            return;        }        int variNum = 0;        int nextK = 0;        int increase = 0;        for (int i =0; i<length; i++) {            if (k<cons[i+1]) {                variNum = i;                nextK = k%cons[i];                increase = k/cons[i];                break;            }        }        int tmp = array[length-variNum-1];        array[length-variNum-1] = array[length-variNum-1+increase];        array[length-variNum-1+increase] = tmp;        for (int i = 0; i< length-variNum; i++) {            xx.append(array[i]);        }        int[] nestArray = new int[variNum];        System.arraycopy(array, length-variNum, nestArray, 0, variNum);        getkIndex(xx, nestArray, nextK, cons);    }}