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[Leetcode]51. N-Queens(C++)

题目描述

题目链接:51. N-Queens

The n-queens puzzle is the problem of placing n queens on an n x 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.

例子

见官网例子。

Constraints

  • 1 <= n <= 9

解题思路

求所有可能解的问题通常都是用回溯法进行暴力遍历加剪枝。这道题也是一样,通过遍历每一行,在遍历某行时,遍历每一列位置,检查该位置对应的列以及对角线是否有棋子,如果没有则添加该棋子,递归地求解下一行。代码如下:

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#include <string>
#include <unordered_set>
#include <vector>

class Solution {
public:
std::vector<std::vector<std::string>> solveNQueens(int n) {
std::vector<std::vector<std::string>> result;
std::unordered_set<int> col_used;
std::unordered_set<int> diag1;
std::unordered_set<int> diag2;
std::vector<std::string> curr;
dfs(0, n, col_used, diag1, diag2, curr, result);
return result;
}

private:
void dfs(int row, int n, std::unordered_set<int>& col_used,
std::unordered_set<int>& diag1, std::unordered_set<int>& diag2,
std::vector<std::string>& curr,
std::vector<std::vector<std::string>>& result) {
if (row == n) {
result.push_back(curr);
return;
}

std::string curr_row(n, '.');
for (int col = 0; col < n; ++col) {
if (col_used.count(col) || diag1.count(col - row) ||
diag2.count(col + row))
continue;
curr_row[col] = 'Q';
col_used.insert(col);
diag1.insert(col - row);
diag2.insert(col + row);
curr.push_back(curr_row);
dfs(row + 1, n, col_used, diag1, diag2, curr, result);
curr.pop_back();
curr_row[col] = '.';
col_used.erase(col);
diag1.erase(col - row);
diag2.erase(col + row);
}
}
};
  • 时间复杂度: O(n!)
  • 空间复杂度: O(n)

GitHub 代码同步地址: 51.NQueens.cpp

其他题目:
GitHub: Leetcode-C++-Solution
博客: Leetcode-Solutions