N-Queens

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.."]
]

这里需要用到回溯（su）算法

回溯算法

1.定义一个解空间，即存放一个解的空间

2.开始遍历，DFS，如果没有一个合适的，回退上一步

这个还要多了解，多谢谢才okay

public class Solution {
    int solution[];
    int num_queen;
    List<String[]>  result = new ArrayList<String[]>();
    
    public List<String[]> solveNQueens(int n) {
        solution = new int[n + 1];
        num_queen = n;
        placeQueen(1);
        return result;
    }
    
    /**放第n个皇后
     * @param n
     */
    public void placeQueen(int n){
        if(n > num_queen){
            String temp[] = new String[num_queen];
            String tempStr = "";
            
            for(int i = 1; i <= num_queen; i++){
                for(int j = 1; j <= num_queen; j++){
                    if(j == solution[i])
                        tempStr += "Q";
                    else
                        tempStr += ".";
                }
                temp[i - 1] = tempStr;
                tempStr = "";
            }
            result.add(temp);
        }else{
            for(int i = 1; i <= num_queen; i++){
                solution[n] = i;                            //将n个皇后放到i列
                if(isValid(n))                                //如果i列有效
                    placeQueen(n + 1);                        //放下一个皇后
            }
        }
    }
    
    /**
     * 第k个皇后的位置是否有效
     * @param k
     * @return
     */
    public boolean isValid(int k){
        for(int i = 1; i < k; i++){
            if(solution[i] == solution[k] || Math.abs(i - k) == Math.abs(solution[i] - solution[k]))
                return false;
        }
        return true;
    }
    
}