【LeetCode】51. N-Queens

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.."]
]

 

这题和Sudoku Solver是一个套路，回溯法尝试所有可能性，将可行解的保存起来。可以对比着看。

由于这里路径尝试本质上是有序的，即1~9逐个尝试，因此无需额外设置状态位记录已经尝试过的方向。

我们先用vector<int>来存放可行解，下标代表行号，元素代表列号。

因此上图中Solution 1用vector<int>表示就是[1,3,0,2]

解题流程就是在下一行中尝试列数(0~n-1)，如果可行则递归下去，如果不可行则弹出继续尝试下一列。

最后将vector<vector<int> > 转为vector<vector<string> >即可。

class Solution {
public:
    vector<vector<string> > solveNQueens(int n) {
        return convert(solve(n), n);
    }
    vector<vector<int> > solve(int n)
    {
        vector<vector<int> > ret;
        vector<int> cur;
        Helper(ret, cur, 0, n);
        return ret;
    }
    void Helper(vector<vector<int> >& ret, vector<int> cur, int pos, int n)
    {
        if(pos == n)
            ret.push_back(cur);
        else
        {
            for(int i = 0; i < n; i ++)
            {
                cur.push_back(i);
                if(check(cur))
                    Helper(ret, cur, pos+1, n);
                cur.pop_back();
            }
        }
    }
    bool check(vector<int> cur)
    {
        int size = cur.size();
        int loc = cur[size-1];
        for(int i = 0; i < size-1; i ++)
        {
            if(cur[i] == loc)
                return false;
            else if(abs(cur[i]-loc) == abs(i-size+1))
                return false;
        }
        return true;
    }
    vector<vector<string> > convert(vector<vector<int> > ret, int n)
    {
        vector<vector<string> > retStr;
        for(int i = 0; i < ret.size(); i ++)
        {
            vector<string> curStr;
            for(int j = 0; j < n; j ++)
            {
                string loc(n, '.');
                loc[ret[i][j]] = 'Q';
                curStr.push_back(loc);
            }
            retStr.push_back(curStr);
        }
        return retStr;
    }
};