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.."] ]
由题意得,这是一个八皇后问题,就是有八个皇后,要放置在8 * 8的棋盘上,其中同一行,同一列,同一个对角线不能存在其他的皇后,问有多少种解法。其实这是一题很经典的深搜题,每一行每一行的放置,然后判断是否合法,一直到8个皇后都放完,表示一种解法,而这样的深搜时间复杂度在O(n!),而判断是否合法则需要O(n),那么总的时间复杂度为O(n! * n),代码如下。
Code(LeetCode运行3ms):
class Solution { public: vector<vector<string>> solveNQueens(int n) { vector<vector<string>> result; vector<int> C(n, -1); dfs(C, result, 0); return result; } void dfs(vector<int>& C, vector<vector<string>>& result, int row) { int n = C.size(); if (row == n) { vector<string> aSolution; for (int i = 0; i < n; i++) { string s(n, '.'); for (int j = 0; j < n; j++) { if (j == C[i]) { s[j] = 'Q'; } } aSolution.push_back(s); } result.push_back(aSolution); return; } for (int j = 0; j < n; j++) { bool valid = isValid(C, row, j); if (!valid) { continue; } C[row] = j; dfs(C, result, row + 1); } } bool isValid(vector<int>& C, int row, int col) { for (int i = 0; i < row; i++) { if (C[i] == col) { return false; } if (abs(i - row) == abs(C[i] - col)) { return false; } } return true; } };