[LeetCode68]Permutations 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, return the kth permutation sequence.

Note: Given n will be between 1 and 9 inclusive.

Analysis:

两个解法。
第一，DFS
递归遍历所有可能，然后累加计算，直至到K为止。

第二，数学解法。

假设有n个元素，第K个permutation是
a1, a2, a3, ..... ..., an
那么a1是哪一个数字呢？

那么这里，我们把a1去掉，那么剩下的permutation为
a2, a3, .... .... an, 共计n-1个元素。 n-1个元素共有(n-1)!组排列，那么这里就可以知道

设变量K1 = K
a1 = K1 / (n-1)!

同理，a2的值可以推导为
a2 = K2 / (n-2)!
K2 = K1 % (n-1)!
.......
a(n-1) = K(n-1) / 1!
K(n-1) = K(n-2) /2!

an = K(n-1)

Java

public String getPermutation(int n, int k) {
	int []num = new int[n];
		int permCount = 1;
		for(int i=0;i<n;i++){
			num[i] = i+1;
			permCount*=(i+1);
		}
		k--;
		StringBuilder target = new StringBuilder();
		for(int i=0;i<n;i++){
			permCount = permCount/(n-i);
			int choosed = k/permCount;
			target.append(num[choosed]);
			for(int j=choosed;j<n-i-1;j++){
				num[j] = num[j+1];
			}
			k = k%permCount;
		}
		return target.toString();
    }

c++

string getPermutation(int n, int k) {
        vector<int>num(n);
		int permCount = 1;
		for(int i=0;i<n;i++){
			num[i] = i+1;
			permCount*=(i+1);
		}
		k--;
		string target;
		for(int i=0;i<n;i++){
			permCount = permCount/(n-i);
			int choosed = k/permCount;
			target.push_back(num[choosed]+'0');
			for(int j=choosed;j<n-i;j++){
				num[j] = num[j+1];
			}
			k = k%permCount;
		}
		return target;
    }