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:
There exist two distinct solutions to the 4-queens puzzle:
[ [".Q..", // Solution 1 "...Q", "Q...", "..Q."], ["..Q.", // Solution 2 "Q...", "...Q", ".Q.."] ]
Analysis:
The classic recursive problem.
1. Use a int vector to store the current state, A[i]=j refers that the ith row and jth column is placed a queen.
2. Valid state: not in the same column, which is A[i]!=A[current], not in the same diagonal direction: abs(A[i]-A[current]) != r-i
3. Recursion:
Start: placeQueen(0,n)
if current ==n then print result
else
for each place less than n,
place queen
if current state is valid, then place next queen place Queen(cur+1,n)
end for
end if
Source Code:
class Solution { public: vector<vector<string> > res; void printres(vector<int> A,int n){ vector<string> r; for(int i=0;i<n;i++){ string str(n,'.'); str[A[i]]='Q'; r.push_back(str); } res.push_back(r); } bool isValid(vector<int> A, int r){ for (int i=0;i<r;i++){ if ( (A[i]==A[r])||(abs(A[i]-A[r])==(r-i))){ return false; } } return true; } void nqueens(vector<int> A, int cur, int n){ if (cur==n){printres(A,n);} else{ for (int i=0;i<n;i++){ A[cur]=i; if (isValid(A,cur)){ nqueens(A,cur+1,n); } } } } vector<vector<string> > solveNQueens(int n) { // Start typing your C/C++ solution below // DO NOT write int main() function res.clear(); vector<int> A(n,-1); nqueens(A,0,n); return res; } };
line #32
ReplyDeletenqueens(A,cur+1,n);
A[cur]=-1; //作者是否缺了这个回溯时的重置-1 吧?