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; } }