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.."] ]
DFS to solve it!
1 class Solution { 2 private: 3 vector<vector<string> > ret; 4 int a[100]; 5 bool canUse[100]; 6 public: 7 bool check(int y, int n) 8 { 9 for(int i = 0; i < n; i++) 10 if (abs(i - n) == abs(y - a[i])) 11 return false; 12 return true; 13 } 14 15 void solve(int dep, int maxDep) 16 { 17 if (dep == maxDep) 18 { 19 vector<string> ans; 20 for(int i = 0; i < maxDep; i++) 21 { 22 string s; 23 for(int j = 0; j < a[i]; j++) 24 s += '.'; 25 s += 'Q'; 26 for(int j = 0; j < maxDep - (a[i] + 1); j++) 27 s += '.'; 28 ans.push_back(s); 29 } 30 ret.push_back(ans); 31 32 return; 33 } 34 35 for(int i = 0; i < maxDep; i++) 36 if (canUse[i] && check(i, dep)) 37 { 38 canUse[i] = false; 39 a[dep] = i; 40 solve(dep + 1, maxDep); 41 canUse[i] = true; 42 } 43 } 44 45 vector<vector<string> > solveNQueens(int n) { 46 // Start typing your C/C++ solution below 47 // DO NOT write int main() function 48 ret.clear(); 49 memset(canUse, true, sizeof(canUse)); 50 solve(0, n); 51 return ret; 52 } 53 };