• 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)
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  • 原文地址:https://www.cnblogs.com/h-hkai/p/9813619.html
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