51. N-Queens java solutions

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.."]
]

这种题目都是使用这个套路，就是用一个循环去枚举当前所有情况，然后把元素加入，递归，再把元素移除，这道题目中不用移除的原因是
我们用一个一维数组去存皇后在对应行的哪一列，因为一行只能有一个皇后，如果二维数组，那么就需要把那一行那一列在递归结束后设回
没有皇后，所以道理是一样的。

public class Solution {
    public List<List<String>> solveNQueens(int n) {
        List<List<String>> ans = new ArrayList<List<String>>();
        recursive(ans,new int[n],0,n);
        return ans;
    }
    
    public void recursive(List<List<String>> ans,int[] columnForRow,int row,int n){
        if(row == n){
            List<String> tmp = new ArrayList<String>();
            for(int i = 0;i < n;i++){
                StringBuilder sb = new StringBuilder();
                for(int j = 0;j < n;j++){
                    if(columnForRow[i] == j)
                        sb.append("Q");
                    else 
                        sb.append(".");
                }
                tmp.add(sb.toString());
            }
            ans.add(tmp);
            return;
        }
        
        for(int i = 0; i < n; i++){
            columnForRow[row] = i;
            if(check(columnForRow,row)){
                recursive(ans,columnForRow,row+1,n);//递归回溯
            }
        }
    }
    
    public boolean check(int[] columnForRow, int row){
        for(int i = 0; i < row; i++){
            if(columnForRow[row] == columnForRow[i] || row-i == Math.abs(columnForRow[row] - columnForRow[i]))
                return false;//检查其他行有列相同或者，对角线有没有元素就是行差和列差不要相同
        }
        return true;
    }
}