# Java ```Java class Solution { private int size; private int count; private void solve(int row, int ld, int rd) { if (row == size) { count++; return; } int pos = size & (~(row | ld | rd)); while (pos != 0) { int p = pos & (-pos); pos -= p; // pos &= pos - 1; solve(row | p, (ld | p) << 1, (rd | p) >> 1); } } public int totalNQueens(int n) { count = 0; size = (1 << n) - 1; solve(0, 0, 0); return count; } } ``` # Python ```Python def totalNQueens(self, n): if n < 1: return [] self.count = 0 self.DFS(n, 0, 0, 0, 0) return self.count def DFS(self, n, row, cols, pie, na): # recursion terminator if row >= n: self.count += 1 return bits = (~(cols | pie | na)) & ((1 << n) — 1) # 得到当前所有的空位 while bits: p = bits & —bits # 取到最低位的1 bits = bits & (bits — 1) # 表示在p位置上放入皇后 self.DFS(n, row + 1, cols | p, (pie | p) << 1, (na | p) >> 1) # 不需要revert cols, pie, na 的状态 ``` # Javascript ```Javascript var totalNQueens = function(n) { let count = 0; void (function dfs(row = 0, cols = 0, xy_diff = 0, xy_sum = 0) { if (row >= n) { count++; return; } // 皇后可以放的地方 let bits = ~(cols | xy_diff | xy_sum) & ((1 << n) - 1); while (bits) { // 保留最低位的 1 let p = bits & -bits; bits &= bits - 1; dfs(row + 1, cols | p, (xy_diff | p) << 1, (xy_sum | p) >> 1); } })(); return count; }; ``` # C/C++ ```C/C++ class Solution { public: int totalNQueens(int n) { dfs(n, 0, 0, 0, 0); return this->res; } void dfs(int n, int row, int col, int ld, int rd) { if (row >= n) { res++; return; } // 将所有能放置 Q 的位置由 0 变成 1，以便进行后续的位遍历 int bits = ~(col | ld | rd) & ((1 << n) - 1); while (bits > 0) { int pick = bits & -bits; // 注: x & -x dfs(n, row + 1, col | pick, (ld | pick) << 1, (rd | pick) >> 1); bits &= bits - 1; // 注: x & (x - 1) } } private: int res = 0; }; ```