[leedcode 51] 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.."]

]

public class Solution {

    List<List<String>> res;

    int A[];

    public List<List<String>> solveNQueens(int n) {

        //经典的DFS，

        //解题思想：借助一个数组储存当前的状态A[i]=j表示第i行，第j列有一个皇后"Q"

        //函数nqueens(int cur,int n),代表cur行之前的已经满足规则，需要从cur到n进行深搜

        //递归过程中，需要验证第cur行是否满足要求，满足要求的规则通过isvalid给出，必须保证前cur行的A[i]!=A[cur],并且不在一个对角线上

        //即，Math.abs(A[i]-A[cur])!=cur-i

        //递归的终止条件是cur==n,此时需要保存中间结果

        //Start: placeQueen(0,n)

        //if current ==n then print result

        //else

        //    for each place less than n,

        //place queen

        //if current state is valid, then place next queen place Queen(cur+1,n)

        //   end for

        //end else

        res=new ArrayList<List<String>>();

        A=new int[n];

        nqueens(0,n);

        return res;

    }

    public void nqueens(int cur,int n){

        if(cur==n) printQ(n);

        else{

            for(int i=0;i<n;i++){

                A[cur]=i;//注意，不需要删除，下次重新赋值

                if(isValid(cur))

                    nqueens(cur+1,n);

            }

        }

    }

    public  void printQ(int n){

        List<String>  ress=new ArrayList<String>();

        for(int i=0;i<n;i++){

            

            StringBuilder seq=new StringBuilder();

            for(int j=0;j<n;j++){

                if(A[i]==j)seq.append("Q");

                else seq.append(".");

            }

            ress.add(seq.toString());

        }

        res.add(ress);

    }

    public boolean isValid(int cur){

        for(int i=0;i<cur;i++){//注意i的范围！

            if(A[i]==A[cur]||Math.abs(A[i]-A[cur])==cur-i)

                    return false;

        }

        return true;

        

        

    }

}