3.30-N-Queens-优快云博客

本文详细阐述了N皇后问题的解决方法，包括算法设计、关键步骤解释以及具体实现过程，通过实例展示了如何使用C++解决该经典问题。

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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.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

简单粗暴的理解，题目的意思就是，给出各个棋子彼此不挨着的所有方案，就是每个棋子旁边必须都有空格~
Answer:

class Solution {
    public:
        typedef vector<string> VS;//vector容器，VS保存string
        typedef vector<VS> VVS;//vector容器，VSS保存VS
        typedef vector<int> VI;//vector容器，VI保存int
        VVS solveNQueens(int n) {
            VVS solutions;//某种方案
            VI solution(n);//第n种方案
            solveNQueensImpl(0, &solution, &solutions);
            return solutions;
        }
    private:
        void solveNQueensImpl(int i, VI* _solution, VVS* _solutions) {
            VI& solution = *_solution;
            VVS& solutions = *_solutions;
            int n = solution.size();
            if (i == n) solutions.push_back(solToStrings(solution));
            else {
                // For each column...
                for (int j = 0; j < n; ++j) {
                    // Skip if there is another queen in this column or diagonals
                    if (isAvailable(solution, i, j)) {
                        solution[i] = j;
                        solveNQueensImpl(i+1, &solution, &solutions);
                    }
                }
            }
        }

        bool isAvailable(const VI& solution, int i, int j) {
            for (int k = 0; k < i; ++k) {
                if (j == solution[k] || abs(i-k) == abs(j-solution[k])) return false;
            }
            return true;
        }

        VS solToStrings(const VI& sol) {
            int n = sol.size();
            VS sol_strings(n);
            for (int i = 0; i < n; ++i) {
                sol_strings[i] = string(n, '.');
                sol_strings[i][sol[i]] = 'Q';
            }
            return sol_strings;
        }
    };