N-Queens--LeetCode

1.题目

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..”]

]

2.题意

在一个N*N的棋盘上放置N个皇后，每行一个并使其不能互相攻击（同一行、同一列、同一斜线上的皇后都会自动攻击）

3.分析

经典的N皇后问题

经典解法为回溯递归，一层层向下扫描

使用pos数组，pos[i]表示第i行皇后所在的列，初始值为-1

从0开始递归，每一行都遍历一次各列，判断如果在该位置放置皇后会不会有冲突

当最后一行皇后放好后，得到一种可能的结果，存入res

4.代码

class Solution {
public:
vector<vector<string>> solveNQueens(int n) {
vector<vector<string>> res;
vector<int> pos(n, -1);
solveNQueensDFS(pos, 0, res);

return res;
}

private:
void solveNQueensDFS(vector<int> &pos, int row, vector<vector<string>> &res)
{
int n = pos.size();
if(row == n)
{
vector<string> out(n, string(n, '.'));
for(int i = 0; i < n; ++i)
out[i][pos[i]] = 'Q';

res.push_back(out);
}
else
{
for(int col = 0; col < n; ++col)
{
if(isValid(pos, row, col))
{
pos[row] = col;
solveNQueensDFS(pos, row + 1, res);
pos[row] = -1;
}
}
}
}

bool isValid(vector<int> &pos, int row, int col)
{
for(int i = 0; i < row; ++i)
{
if(pos[i] == col || abs(row - i) == abs(col - pos[i]))
return false;
}

return true;
}
};