[leetcode]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.."]
]

算法思路：
回溯算法，经典的八皇后问题，不多啰嗦，不明白的请戳这里；

public class Solution {
List<String[]> result = new ArrayList<String[]>();
    public List<String[]> solveNQueens(int n) {
        if(n <= 0) return result;
        int[] hash = new int[n];
        dfs(hash,0,n);
        return result;
    }
    private void dfs(int[] hash,int row,int n){
        if(row == n ){
            String[] node = new String[n];
            for(int i = 0; i < n; i++){
                StringBuilder sb = new StringBuilder();
                for(int j = 0; j < n;sb.append('.'),j++);
                sb.setCharAt(hash[i], 'Q');
                node[i] = sb.toString();
            }
            result.add(node);
        }
        for(int i = 0; i < n; i++){
            if(!isConflict(hash, row, i)){
                hash[row] = i;
                dfs(hash, row + 1, n);
                hash[row] = 0;//track back
            }
        }
    }
    private boolean isConflict(int[] hash,int row,int column){
        for(int i = 0; i < row; i++){
            if(hash[i] == column) return true;
            else if(i - hash[i] == row - column || hash[i] + i == row + column) return true;
        }
        return false;
    }
}