[LintCode] N-Queens
N-Queens
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.
Example
There exist two distinct solutions to the 4-queens puzzle:
[
[".Q..", // Solution 1
"...Q",
"Q...",
"..Q."],
["..Q.", // Solution 2
"Q...",
"...Q",
".Q.."]
]
1 class Solution { 2 public: 3 /** 4 * Get all distinct N-Queen solutions 5 * @param n: The number of queens 6 * @return: All distinct solutions 7 * For example, A string '...Q' shows a queen on forth position 8 */ 9 bool isOK(vector<string> &v, int n, int x, int y) { 10 for (int i = 0; i < x; ++i) 11 if (v[i][y] == 'Q') return false; 12 for (int i = 1; x - i >= 0 && y - i >= 0; ++i) 13 if (v[x-i][y-i] == 'Q') return false; 14 for (int i = 1; x - i >= 0 && y + i < n; ++i) 15 if (v[x-i][y+i] == 'Q') return false; 16 return true; 17 } 18 void dfs(vector<vector<string>> &res, vector<string> &v, int n, int idx) { 19 if (idx == n) { 20 res.push_back(v); 21 return; 22 } 23 for (int i = 0; i < n; ++i) { 24 v[idx][i] = 'Q'; 25 if (isOK(v, n, idx, i)) dfs(res, v, n, idx + 1); 26 v[idx][i] = '.'; 27 } 28 } 29 vector<vector<string> > solveNQueens(int n) { 30 // write your code here 31 vector<vector<string>> res; 32 string s; 33 for (int i = 0; i < n; ++i) s.push_back('.'); 34 vector<string> v(n, s); 35 dfs(res, v, n, 0); 36 return res; 37 } 38 };
