60. Permutation Sequence
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The set [1,2,3,…,n]
contains a total ofn! 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.
思路: 将元素按从小到大的顺序保存在list中,从第一个开始判断应该是什么数字,依次根据下标从list中选取数据,然后删除已用的数据.
code:
public class Solution {
public String getPermutation(int n, int k) {
StringBuilder res=new StringBuilder("");
int all=1;
int temp=n-1;
List<Integer> list=new ArrayList<Integer>();
for(int i=1;i<=n;i++)
list.add(i);
while(temp>1)
all*=(temp--);
for(int i=0;i<n-1;i++){
temp=(k-1)/all;// 注意这里是k-1
res.append(list.get(temp));
list.remove(temp);
k-=temp*all;
all/=(n-i-1);
}
res.append(list.get(0));
return res.toString();
}
}
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