題目

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.."]
]

解題之法

class Solution {
public:
    vector<vector<string> > solveNQueens(int n) {
        vector<vector<string> > res;
        vector<int> pos(n, -1);
        solveNQueensDFS(pos, 0, res);
        return res;
    }
    void solveNQueensDFS(vector<int> &pos, int row, vector<vector<string> > &res) {
        int n = pos.size();
        if (row == n) {
            vector<string> out(n, string(n, '.'));
            for (int i = 0; i < n; ++i) {
                out[i][pos[i]] = 'Q';
            }
            res.push_back(out);
        } else {
            for (int col = 0; col < n; ++col) {
                if (isValid(pos, row ,col)) {
                    pos[row] = col;
                    solveNQueensDFS(pos, row + 1, res);
                    pos[row] = -1;
                }
            }
        }
    }
    bool isValid(vector<int> &pos, int row, int col) {
        for (int i = 0; i < row; ++i) {
            if (col == pos[i] || abs(row - i) == abs(col - pos[i])) {
                return false;
            }
        }
        return true;
    }
};

分析

經(jīng)典的N皇后問題，基本所有的算法書中都會包含的問題，經(jīng)典解法為回溯遞歸，一層一層的向下掃描，需要用到一個pos數(shù)組，其中pos[i]表示第i行皇后的位置，初始化為-1，然后從第0開始遞歸，每一行都一次遍歷各列，判斷如果在該位置放置皇后會不會有沖突，以此類推，當?shù)阶詈笠恍械幕屎蠓藕煤?，一種解法就生成了，將其存入結(jié)果res中，然后再還會繼續(xù)完成搜索所有的情況，