lintcode-medium-N Queens II-优快云博客

本文介绍了一种解决N皇后问题的方法，该方法不仅找出所有可能的棋盘配置，还计算了对于给定的N值有多少种不同的解决方案。通过递归回溯的方式，文章中的代码实现了对N皇后问题解决方案总数的计算。

Follow up for N-Queens problem.

Now, instead outputting board configurations, return the total number of distinct solutions.

Example

For n=4, there are 2 distinct solutions.

class Solution {
    /**
     * Calculate the total number of distinct N-Queen solutions.
     * @param n: The number of queens.
     * @return: The total number of distinct solutions.
     */
    public int totalNQueens(int n) {
        //write your code here
        
        ArrayList<ArrayList<String>> result = new ArrayList<ArrayList<String>>();
        
        if(n <= 0) 
            return 0;
        
        int[] position = new int[n];
        
        helper(result, 0, position, n);
        
        return result.size();
    }
    
        public void helper(ArrayList<ArrayList<String>> result, int row, int[] position, int n){
        
        if(row == n){
            ArrayList<String> solution = new ArrayList<String>();
            for(int i = 0; i < n; i++){
                StringBuilder line = new StringBuilder();
                
                for(int j = 0; j < n; j++){
                    if(j == position[i])
                        line.append('Q');
                    else
                        line.append('.');
                }
                
                solution.add(line.toString());
            }
            
            result.add(new ArrayList<String>(solution));
            return;
        }
        
        for(int i = 0; i < n; i++){
            position[row] = i;
            if(valid(position, row, n))
                helper(result, row + 1, position, n);
        }
        
        return;
    }
    
    public boolean valid(int[] position, int row, int n){
        
        for(int i = row - 1; i >= 0; i--){
            if(position[row] == position[i] || Math.abs(position[row] - position[i]) == row - i)
                return false;
        }
        
        return true;
    }
};