LeetCode——N-Queens

Description：

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.."]
]

经典的n皇后问题，有一个经典的回溯法。具体请参考：N-Queens

在这个问题中需要用List来记录sum个解矩阵，并返回。

代码：

public class Solution {

    public int[] x; //当前解
    public int sum; //解的个数
    public int n; //皇后个数
    public List<List<String>> resList;

    public List<List<String>> solveNQueens(int n) {
        this.resList = new ArrayList<List<String>>();

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

        }

        this.x = new int[n + 1];

        for(int i=0; i<=n; i++) {
            this.x[i] = 0;
        }

        this.sum = 0;
        this.n = n;

        backTrack(1);

        return resList;
    }

    public void backTrack(int t) {
        if(t > n) {
            List<String> pRes = new ArrayList<String>();
            for(int i=1; i<=n; i++) {
                StringBuilder row = new StringBuilder();
                for(int j=1; j<=n; j++) {
                    if(this.x[i] == j) {
                        row.append("Q");
                    }
                    else {
                        row.append(".");
                    }
                }
                pRes.add(row.toString());

            }
            resList.add(pRes);
            sum ++;
        }
        else {
            for(int i=1; i<=n; i++) {
                this.x[t] = i;
                if(place(t)) {
                    backTrack(t + 1); //回溯
                }
            }
        }
    }
    /**
     * 判断当前位置是否合法
     */
    public boolean place(int k) {
        for(int j=1; j<k; j++) {
            if(Math.abs(j-k) == Math.abs(this.x[j]-this.x[k]) ||
                    this.x[j] == this.x[k]) {
                return false;
            }
        }
        return true;
    }

}