Leetcode - Queens I && II

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.."]
]

[Thoughts]： Queens数组，index为行号，value是改行放queen的列号

public class Solution {
    public List<String[]> solveNQueens(int n) {
        List<String[]> res = new ArrayList<String[]>();
        int[] q = new int[n];
        placeQ(res, q, 0);
        return res;
    }
    
    public void placeQ(List<String[]> res, int[] q, int row){
        if(row==q.length){
            addResult(res, q);
            return;
        }
        for(int i=0; i<q.length; i++){
            if(isValid(q, row, i)){
                q[row] = i;
                placeQ(res, q, row+1);
            }
        }
    }
    
    public boolean isValid(int[] q, int row, int col){
        for(int i=0; i<row; i++){
            if(q[i] == col) 
                return false;
            if(Math.abs(i-row) == Math.abs(q[i]-col))
                return false;
        }
        return true;
    }
    
    public void addResult(List<String[]> res, int[] q){
        String[] line = new String[q.length];
        for(int i=0; i<q.length; i++){
            StringBuilder sb = new StringBuilder();
            for(int j=0; j<q.length; j++){
                sb.append(q[i]==j ? "Q" : ".");
            }
            line[i] = sb.toString();
        }
        res.add(line);
    }
}

II: Follow up for N-Queens problem.
Now, instead outputting board configurations, return the total number of distinct solutions.
和I的思路一样，区别就是不用去拼写string，设置一个全局变量count去数个数就行了。

public class Solution {
    private int count;
    public int totalNQueens(int n) {
        if(n<=0) return -1;
        int[] queens = new int[n];
        placeQueens(queens, 0);
        return count;
    }
    
    public void placeQueens(int[] queens, int row){
        //only difference with I, 就是处理solution的方式，此题计数就可以了。
        if(row==queens.length){
            count++;
            return;
        }
        
        for(int i=0; i<queens.length; i++){
            if(isValid(queens, row, i)){
                queens[row] = i;
                placeQueens(queens, row+1);
            }
        }
    }
    
    public boolean isValid(int[] queens, int row, int col){
        for(int i=0; i<row; i++){
            if(queens[i]==col)
                return false;
            if(Math.abs(row-i) == Math.abs(col-queens[i]))
                return false;
        }
        return true;
    }
}