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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