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leetcode 刷题之路 95 N-Queens I

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

N皇后问题的变种,要求直接输出N皇后的解法数目。这道题可以在N-Queens I的基础上增加计数功能,在每求得一个成功的解时(行数为N时)使计数变量递增即可。题目不要求输出具体的解法,因此可以做一点优化,使用position数组用来表示皇后的位置,position[i]的值表示第i行的皇后所处的列,这样子省却字符串数组的开销同时判断是否合法的isValid函数也更加简洁。

Accepted Solution:

 class Solution {
 public:
	 int totalNQueens(int n)
	 {
		 int res = 0;
		 vector<int> position(n, -1);
		 helper(res, position, 0, n);
		 return res;
	 }
	 void helper(int &res, vector<int> &position, int row, int n)
	 {
		 if (row == n)
			 res++;
		 for (int clmn = 0; clmn<n; clmn++)
		 {
			 if (isValid(position, row, clmn, n))
			 {
				 position[row] = clmn;
				 helper(res, position, row + 1, n);
				 position[row] = -1;
			 }
		 }
	 }
	 bool isValid(vector<int> &position, int row, int clmn, int n)
	 {
		 for (int i = 0; i<n; i++)
		 if (position[i] == clmn || (position[i] != -1 && abs(row - i) == abs(clmn - position[i])))
		 {
			 return false;
		 }
			 
		 return true;
	 }
 };