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

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

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++

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;
};