51. N-Queens (Graph; WFS)

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.."]
]

思路：WFS，按行递归，在每个递归过程中按列循环，此时可能会有多种选择，所以使用回溯法

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        result.clear();
        string str="";
        for(int i = 0; i<n;i++) str+= ".";
        vector<string> answer(n,str);
        vector<bool> columnFlag(n,false);
        dfs(answer,columnFlag, n, 0);
        return result;
    }
    void dfs(vector<string> &answer, vector<bool> &columnFlag, int n, int depth){ //depth表示行号
        if(depth == n) //递归结束条件
        {
            result.push_back(answer);
            return;
        }

        bool haveAnswer = true;
        for(int i = 0; i < n; i++) //按列循环搜索
        {
            if(columnFlag[i]) continue; //check column

            for (int j = 1; depth-j >= 0 && i-j >= 0; j++) //check upper left diagonal
            {
                if(answer[depth-j][i-j]==‘Q‘)
                {
                    haveAnswer = false;
                    break;
                }
            }
            if(!haveAnswer)
            {
                haveAnswer = true;
                continue;
            }

            for (int j = 1; depth-j >= 0 && i+j <= n-1; j++)  //check upper right diagonal
            {
                if(answer[depth-j][i+j]==‘Q‘)
                {
                    haveAnswer = false;
                    break;
                }
            }
            if(!haveAnswer)
            {
                haveAnswer = true;
                continue;
            }

            //找到一种answer，递归
            answer[depth][i] = ‘Q‘;
            columnFlag[i] = true;
            dfs(answer, columnFlag, n, depth+1);

            //回溯
            answer[depth][i] = ‘.‘;
            columnFlag[i] = false;
        }
    }
private:
    vector<vector<string>> result;
};