LeetCode 60: Permutation Sequence

LeetCode 60: Permutation Sequence

算法思路

已知以每个数字开头的序列个数为(n - 1)!，根据(k - 1) / (n - 1) ! + 1可以知道第一个数字是几，然后递归，求出之后的数字；

算法中数组flag记录数字是否被使用，factorials记录阶乘值，如果在递归函数中直接求阶乘会导致超时；

每次递归n的值会减1，递归终止的条件即n为0，此外k的值变为k - position * factorials[n - 1];

代码

public class Solution {

public String ret = "";
public String getPermutation(int n, int k) {

boolean[] flag = new boolean
; // 0 ~ n - 1
//all haven't been used
for (int i = 0; i < flag.length; i++) flag[i] = false;

int[] factorials = new int[n + 1];
factorials[0] = 1;
for (int i = 1; i <= n; i++) factorials[i] = factorials[i - 1] * i;

getPermutation(n, k, flag, factorials);
return ret;
}
public void getPermutation(int n, int k, boolean[] flag, int[] factorials) {

if (n == 0) return;
int factorial = factorials[n - 1];
int position = (k - 1) / factorial;

int count = -1, i;
for (i = 0; i < flag.length; i++) {
if (!flag[i]) count++;
if (count == position) break;
}
flag[i] = true;
ret = ret + (i + 1);

getPermutation(n - 1, k - position * factorial, flag, factorials);
}
}