### leetcode Question 59: N-Queens

N-Queens

The n-queens puzzle is the problem of placing n queens on an nÃ—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.."]
]


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;
}
};


#### 1 comment:

1. line #32
nqueens(A,cur+1,n);
A[cur]=-1; //作者是否缺了这个回溯时的重置-1 吧？