[LeetCode] N-Queens

最新推荐文章于 2020-07-10 00:48:29 发布

1.02^365的成长裂变

最新推荐文章于 2020-07-10 00:48:29 发布

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分类专栏： LeetCode1

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本文介绍了一种解决N皇后问题的有效算法。通过递归搜索和冲突检查的方式，在N×N的棋盘上放置N个皇后，确保任意两个皇后都不会互相攻击。文章提供了具体的代码实现，展示了如何找到所有可能的解决方案。

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[Problem]

The n-queens puzzle is the problem of placing n queens on an n?n chessboard such that no two queens attack each other.

[LeetCode] N-Queens - coder007 - Coder007的博客

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.."]
]

[Solution]

class Solution {
public:
    // is the new Queen valid
    bool valid(int row, int column, vector<pair<int, int> > &located){
        for(int i = 0; i < located.size(); ++i){
            if(located[i].first + located[i].second == row + column || located[i].second - located[i].first == column - row){
                return false;
            }
        }
        return true;
    }
   
    // search location
    vector<vector<string> > search(bool column[], int n, int mth, vector<pair<int, int> > &located){
        vector<vector<string> > res;
   
        // the last row
        if(mth == n - 1){
            for(int i = 0; i < n; ++i){
                if(column[i] == false && valid(mth, i, located)){
                    string str(n, '.');
                    str[i] = 'Q';
   
                    // add result
                    vector<string> tmp;
                    tmp.push_back(str);
                    res.push_back(tmp);
                }
            }
        }
        else{
            for(int i = 0; i < n; ++i){
                // valid in the ith column
                if(column[i] == false && valid(mth, i, located)){
   
                    // locate a Queen in (mth, i)
                    column[i] = true;
                    located.push_back(make_pair(mth, i));
                    string str(n, '.');
                    str[i] = 'Q';
   
                    // search in the next row
                    vector<vector<string> > r = search(column, n, mth+1, located);
                    for(int j = 0; j < r.size(); ++j){
                        r[j].insert(r[j].begin(), str);
                        res.push_back(r[j]);
                    }
   
                    // back
                    column[i] = false;
                    located.pop_back();
                }
            }
        }
        return res;
    }
   
    // n queen
    vector<vector<string> > solveNQueens(int n) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
       
        // the columns with Queen
        bool column[n];
        memset(column, false, sizeof(column));
   
        // located Queens
        vector<pair<int, int> > located;
   
        // search
        return search(column, n, 0, located);
    }
};