LeetCode:N-Queens

N-Queens

Total Accepted: 55554  Total Submissions: 212496  Difficulty: Hard

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.."]
]

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  (H) N-Queens II

思路：

将N-Queens II中的解做一个转化即可。

c++ code:

class Solution {
public:
    vector<vector<string>> solveNQueens(int n) {
        
        vector<vector<string>> res;
        vector<int> nums(n);
        for(int i=0;i<n;i++)
            nums[i]=i;
        
        permute(res, nums,0);
        return res;
    }
    
    // 自定义函数
    void permute(vector<vector<string>> &res, vector<int> &nums, int pos) {
        
        int n = nums.size();
        if(pos==n) {
            if(check(nums)) {
                vector<string> ans;
                for(int i=0;i<n;i++) {
                    string str(n, '.');
                    str[nums[i]] = 'Q';
                    ans.push_back(str);
                }
                res.push_back(ans);
            }
            return;
        }
        for(int i=pos;i<n;i++) {
            swap(nums[pos], nums[i]);
            permute(res, nums, pos+1);
            swap(nums[i], nums[pos]);
        }
          
    }
      
    bool check(vector<int> &nums) {
        
        int n = nums.size();
        for(int i = 0;i < n;i++) {
            for(int j = i + 1;j < n;j++)
                // 判断是否在主、副对角线上
                if(i-j == nums[i]-nums[j] || j-i == nums[i]-nums[j])
                    return false;
        }
        return true;
    }
    
};