(java)N-Queens

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 all distinct solutions to the n-queens puzzle.

Each solution contains a distinct board configuration of the n-queens' placement, where 'Q' and '.' both indicate a queen and an empty space respectively.

For example,
There exist two distinct solutions to the 4-queens puzzle:

[ [".Q..",  // Solution 1  "...Q",  "Q...",  "..Q."], ["..Q.",  // Solution 2  "Q...",  "...Q",  ".Q.."]]

思路:n皇后问题，回溯算法的最经典问题，回溯探索就行了，太简单，太经典，就直接贴代码了

代码如下(已通过leetcode)

public class Solution {
List<List<String>> lists=new ArrayList<List<String>>();
   public List<List<String>> solveNQueens(int n) {
    char[][] queens = new char[n][n];
       for(int i = 0;i < n; i++){
           for(int j = 0;j < n; j++){
               queens[i][j] = '.';
           }
       }
       genaratenqueens(queens, n,0);
    return lists;
   
   }
   
   


   private void genaratenqueens(char[][] queens, int n, int row) {
// TODO Auto-generated method stub
if(row==n) {
List<String> temp=new ArrayList<String>();
for(int i=0;i<n;i++) {
temp.add(String.valueOf(queens[i]));

}
lists.add(temp);
}
for(int i=0;i<n;i++) {
if(canput(row, i, n, queens)) {
queens[row][i]='Q';
genaratenqueens(queens, n, row+1);
queens[row][i]='.';
}
}
}






public boolean canput(int x,int y,int n,char[][] queens){
       int i;
       int j;
       for(i=0;i<n;i++) {
        for(j=0;j<n;j++) {
        if(Math.abs(i-x)==Math.abs(j-y)) {
        if(queens[i][j]=='Q') return false;
        }
        }
       }
       i=0;
       for(i=0;i<n;i++) {
        if(queens[i][y]=='Q') return false;
       }
       return true;
   }
}