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 th, return the kth permutation sequence.
Note: Given n will be between 1 and 9 inclusive.
我刚开始尝试着把，所有的全排列都列出来，然后，找到对应位置的内容，但是这样效率太低，直接超时了，上网差了一下，和数学有关系。。。这题打算先放一放，放一下别人的代码吧。
public String getPermutation(int n, int k) {
char[] nums = new char[]{'1','2','3','4','5','6','7','8','9'};
String tmp = "";
for(int i=0;i<n;i++) {
tmp += nums[i];
}
StringBuffer s = new StringBuffer(tmp);
String r = "";
while(k>0&&!s.toString().equals("")) {
// 计算 (n-1)的排列个数cnt
int cnt = 1, i = s.length()-1;
while(i > 1) {
cnt*=i;
i-=1;
}
int pos = (k-1)/cnt;
r += s.charAt(pos);
s = s.deleteCharAt(pos);
k -= pos * cnt;
}
return r;
}
