51. 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:
Input: 4 Output: [ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ] Explanation: There exist two distinct solutions to the 4-queens puzzle as shown above.
AC code:
class Solution { public: vector<vector<string>> solveNQueens(int n) { int res = 0; vector<vector<string>> v; vector<string> nqueens(n, string(n, '.')); solve(v, nqueens, n, 0); return v; } void solve(vector<vector<string>>& v, vector<string>& nqueens, int n, int row) { if (row == n) { v.push_back(nqueens); return; } for (int col = 0; col < n; ++col) { if (judge(nqueens, n, row, col)) { nqueens[row][col] = 'Q'; solve(v, nqueens, n, row+1); nqueens[row][col] = '.'; } } } bool judge(vector<string> nqueens, int n, int x, int y) { // up and down for (int i = 0; i < n && i != x; ++i) { if (nqueens[i][y] == 'Q') return false; } // right and left for (int i = 0; i < n && i != y; ++i) { if (nqueens[x][i] == 'Q') return false; } // left up for (int i = x-1, j = y-1; i >= 0 && j >= 0; --i, --j) { if (nqueens[i][j] == 'Q') return false; } // right up for (int i = x-1, j = y+1; i >= 0 && j < n; --i, ++j) { if (nqueens[i][j] == 'Q') return false; } return true; } };
Runtime: 32 ms, faster than 8.09% of C++ online submissions for N-Queens.
永远渴望,大智若愚(stay hungry, stay foolish)