N-Queens II——Leetcode

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

计算N皇后的合法解的数量。

解题思路：这次使用了更优化的方式判断棋盘上的冲突，abs(curr.row-row)==abs(curr.col-col)

board数组纪录每行的列的位置，如board[1]=3，代表第一行第三列上有皇后。

public class Solution {
    
    int[] board;
    int res = 0;
    public int totalNQueens(int n) {
        board = new int[n];
        helper(0,n);
        return res;
    }
    
    private void helper(int k,int n){
        if(k>=n){
            res++;
            return;
        }
        for(int i=0;i<n;i++){
            if(valid(k,i)){
                board[k]=i;
                helper(k+1,n);
            }
        }
        return ;
    } 
    
    private boolean valid(int x, int y){
        int row = 0;
        while(row<x){
            if(board[row]==y||Math.abs(x-row)==Math.abs(y-board[row])){
                return false;
            }
            row++;
        }
        return true;
    }
}