Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
class Solution {
public:
void backtrace(int &ans, vector<int> &board, int k, int n) {
if(k >= n - 1){
ans++;
return;
}
k++;
board[k] = -1;
for(int t = 0; t < n; t++) {
int p;
for(p = 0; p < k; p++){
if(board[p] == t || k - p == t - board[p] || k - p == board[p] - t){
break;
}
}
if(p == k){
board[k] = t;
backtrace(ans, board, k, n);
board[k] = -1;
}
}
}
int totalNQueens(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
int ans = 0;
vector<vector<char> > ban;
for(int i = 0; i < n; i++){
vector<char> b(n, 0);
ban.push_back(b);
}
vector<int> board(n, -1);
backtrace(ans, board, -1, n);
return ans;
}
};
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