LeetCode_60 Permutation Sequence

Link to original problem: 这里写链接内容

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.

Related problem:

31 Next Permutation: 这里写链接内容

46 Permutations: 这里写链接内容

因为这题只需要返回第k个permutation的结果就行了，所以没必要采取全遍历的算法，求出所有结果，再遍历一次得到第k个String。我们只需计算使用k计算每个位置该出现什么字符即可。具体代码如下：

public class Solution {
public String getPermutation(int n, int k) {
if(n == 1) return "1";
int[] mask = new int
;
mask[0] = 1;
mask[1] = 1;
for(int ii = 2; ii < n; ii++){
mask[ii] = mask[ii-1]*ii;
}
k--;
boolean[] used = new boolean
;
char[] res = new char
;
for(int ii = 0; ii < n; ii++){
int which = k/mask[n-1-ii];
int toWhich = -1;
int index = -1;
while(index < n && toWhich < which){
if(used[++index] == false) toWhich++;
}
res[ii] = (char) (index + '1');
used[index] = true;
k = k%mask[n-1-ii];
}
return new String(res);
}
}