LeetCode 60. Permutation Sequence

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):

  1. "123"
  2. "132"
  3. "213"
  4. "231"
  5. "312"
  6. "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;i0&&!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;
        }

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