【N-Queens】cpp

题目：

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.."]
]

代码：

class Solution {
public:
        static vector<vector<string> > solveNQueens(int n)
        {
            vector<vector<string> > ret;
            if ( n==0 ) return ret;
            vector<string> tmp;
            vector<bool> colUsed(n,false);
            vector<bool> diagUsed1(2*n-1,false);
            vector<bool> diagUsed2(2*n-1,false);
            Solution::dfs(ret, tmp, n, colUsed, diagUsed1, diagUsed2);
            return ret;
        }
        static void dfs(
            vector<vector<string> >& ret,
            vector<string>& tmp,
            int n,
            vector<bool>& colUsed,
            vector<bool>& diagUsed1,
            vector<bool>& diagUsed2 )
        {
            const int row = tmp.size();
            if ( row==n )
            {
                ret.push_back(tmp);
                return;
            }
            string curr(n,‘.‘);
            for ( size_t col = 0; col<n; ++col )
            {
                if ( !colUsed[col] && !diagUsed1[col+n-1-row] && !diagUsed2[col+row] )
                {
                    colUsed[col] = !colUsed[col];
                    diagUsed1[col+n-1-row] = !diagUsed1[col+n-1-row];
                    diagUsed2[col+row] = !diagUsed2[col+row];
                    curr[col] = ‘Q‘;
                    tmp.push_back(curr);
                    Solution::dfs(ret, tmp, n, colUsed, diagUsed1, diagUsed2);
                    tmp.pop_back();
                    curr[col] = ‘.‘;
                    diagUsed2[col+row] = !diagUsed2[col+row];
                    diagUsed1[col+n-1-row] = !diagUsed1[col+n-1-row];
                    colUsed[col] = !colUsed[col];
                }
            }
        }
};

tips：

深搜写法：

1. 找到一个解的条件是tmp的长度等于n

2. 在一列中遍历每个位置，是否能够放置Q，并继续dfs；返回结果后，回溯tmp之前的状态，继续dfs。

一开始遗漏了对角线也不能在有超过一个Q的条件，补上之后就AC了。