[leetcode] Permutation Sequence

最新推荐文章于 2019-05-17 11:20:47 发布

原创最新推荐文章于 2019-05-17 11:20:47 发布 · 669 阅读

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本文介绍了一种算法，用于找出由1到n整数构成的所有排列中的第k个排列序列。通过递归或数学组合的方法，高效地解决了这一问题，并提供了两种实现方案。

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 k^th permutation sequence.

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

class Solution {
public:
    string getPermutation(int n, int k) {
		// Start typing your C/C++ solution below
		// DO NOT write int main() function
		int count=0;
		int *A=new int[n];
		for(int i=0 ; i<n ; i++)
			A[i]=i+1;
		vector<string> str(k);
		_permutations(A,n,0,count,k,str);
		return str[k-1];
	}
	void _permutations(int A[] , int n , int depth , int& count , int k, vector<string>& str){
		if(depth==n-1){
			string temp = convert(A,n);
			if(count<k){
				str[count]=temp;
				count++;
				if(count==k)
					sort(str.begin(),str.end());
			}
			else
				Insert(str,temp);
		}
		for(int i=depth ; i<n ; i++){
			swap(A[depth],A[i]);
			_permutations(A,n,depth+1,count,k,str);
			swap(A[depth],A[i]);
		}
	}
	string convert(int A[] , int n){
		string ret(n,'\0');
		for(int i=0 ; i<n ;i++){
			ret[i]='0'+A[i];
		}
		return ret;
	}
	void Insert(vector<string>& pstr , string str){
		int count=pstr.size()-1;
		if(str>pstr[count])
			return ;
		pstr[count]=str;
		for(int i=count-1 ; i>=0 ; i--){
			if(pstr[i]>str)
				swap(pstr[i],pstr[i+1]);
		}
		return ;
	}
};

看一看神一样的代码：

class Solution {
public:
    string getPermutation(int n, int k) {
        string str;
        vector<int> factorial(n + 1, 1);
        for (int i = 1; i <= n; ++i) {
            str += i + '0';
            factorial[i] = factorial[i-1] * i;
        }
        string perm;
        --k;    // convert to 0-based index
        for (int i = n - 1; i >= 0; --i) {
            int quotient = k / factorial[i];
            perm += str[quotient];
            str.erase(quotient, 1);
            k %= factorial[i];
        }
        return perm;
    }
};