LeetCode 52. N-Queens II 回溯法+约束编程 C++

The n-queens puzzle is the problem of placing n queens on an n×n chessboard such that no two queens attack each other.
Given an integer n, return the number of distinct solutions to the n-queens puzzle.
n 皇后问题研究的是如何将 n 个皇后放置在 n×n 的棋盘上，并且使皇后彼此之间不能相互攻击。
给定一个整数 n，返回 n 皇后不同的解决方案的数量。

Example:

Input: 4
Output: 2
Explanation: There are two distinct solutions to the 4-queens puzzle as shown below.
[
 [".Q..",  // Solution 1
  "...Q",
  "Q...",
  "..Q."],

 ["..Q.",  // Solution 2
  "Q...",
  "...Q",
  ".Q.."]
]

C++

//algorithm: backtrack
class Solution {
public:
    int totalNQueens(int n) {
		board = vector<string>(n,string(n,'.'));
		cols = vector<int>(n,0);
		diag1 = vector<int>(2 * n - 1, 0);
		diag2 = vector<int>(2 * n - 1, 0);
		
		backtrack(n, 0);
		return cnt;
	}
private:
	vector<string> board;
	vector<int> cols;
	vector<int> diag1;
	vector<int> diag2;
	int cnt;
	bool available(int x, int y, int n){
		return !cols[x] 
		    && !diag1 [x+y] 
			&& !diag2[x - y + n -1];
	}
	void update(int x, int y, int n, bool isput){
		cols[x] = isput;
		diag1[x+y] = isput;
		diag2[x - y + n -1] = isput;
		
		board[x][y] = isput ? 'Q' : '.'; 
	}
	//Try to put the queen on y-th row
	void backtrack(int n, int y){
		if(y == n){
			cnt++;
			return;
		}
		//try evert column 
		for(int x = 0; x < n; x++){
			if(!available(x,y,n)) continue;
			update(x,y,n,true);
			backtrack(n,y+1);
			update(x,y,n,false);
		}
	}
};

C++ ②

//algorithm: backtrack
class Solution {
public:
    int cnt;
    vector<int> cols;
    vector<int> diag1;
    vector<int> diag2;
    vector<string>board;
    
    int totalNQueens(int n) {
        cnt = 0;
    	cols = vector<int> (n, 0);
    	diag1 = vector<int> (2*n-1, 0);
    	diag2 = vector<int> (2*n-1, 0);
    	board = vector<string> (n,string(n,'.'));
    	backtrack(n,0);
    	return cnt;
	}
    bool available(int x, int  y, int n){
		return !cols[x] && !diag1[x+y] && !diag2[x - y + n - 1];
	}
	void update(int x, int y, int n, bool isput){
		cols[x] = isput;
		diag1[x+y] = isput;
		diag2[x-y+n-1] = isput;
		
		board[x][y] = isput ? 'Q' : '.';
	}
	void backtrack(int n ,int y){
		if(y == n){
			cnt++;
			return;
		}
		for(int x = 0; x < n; x++){
			if(!available(x,y,n)) continue;
			update(x, y, n, true);
			backtrack(n,y+1);
			update(x, y, n, false);
		}
	}
};

小结：是前一题的子问题LeetCode 51. N-Queens，直接copy上一题的代码，最后输出cnt即可。