题目:
The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return all distinct solutions to the n-queens puzzle.
Each solution contains a distinct board configuration of the n-queens' placement, where 'Q'
and '.'
both indicate a queen and an empty space respectively.
For example,
There exist two distinct solutions to the 4-queens puzzle:
[ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ]
分析与解答:递归的DFS,不算难,注意判断是否合理的条件。
class Solution{ public: vector<vector<string> > result; vector<vector<string> > solveNQueens(int n){ vector<string> path(n,string(n,'.')); dfs(0,0,n,path); return result; } void dfs(int row,int col,int n,vector<string> &path){ if(row == n){ result.push_back(path); return; } for(int j = 0;j != n;++j){ path[row][j] = 'Q'; if(isSafe(path,n,row,j)){ dfs(row+1,0,n,path); } path[row][j] = '.'; } return; } bool isSafe(vector<string> &path,int n,int row,int col){ for(int i = 0;i != row;++i){ if(path[i][col] == 'Q'){ return false; } } int i = row - 1, j = col - 1, k = col + 1; while((i < row&&i >=0) &&(j >= 0 ||k < n)){ if(path[i][j--] == 'Q'|| path[i][k++] == 'Q'){ return false; } i--; } return true; } };