LeetCode 52. N-Queens II

原题链接在这里：https://leetcode.com/problems/n-queens-ii/

题目：

Follow up for N-Queens problem.

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

题解：

Here it needs to return possible count. We could use an array of length 1 to store accumlated count.

因为Java是pass by value, 需要这类object结构来保留中间结果.

Time Complexity: expontenial.

Space: O(n). 生成的row大小为n. 用了O(n)层stack.

AC Java:

public class Solution {
    public int totalNQueens(int n) {
        int [] res = new int[1];
        dfs(n,0,new int[n], res);
        return res[0];
    }
    //递归写dfs, 每次往row里面加一个数，若是合法就继续dfs
    private void dfs(int n, int cur, int [] row, int [] res){
        if(cur == n){
            res[0]++;
            return;
        }
        for(int i = 0; i<n; i++){
            row[cur] = i;
            if(isValid(cur, row)){
                dfs(n, cur+1, row, res);
            }
        }
    }
    //检查现在row 还是否合法
    private boolean isValid(int cur, int [] row){
        for(int i = 0; i<cur; i++){
            if(row[cur] == row[i] || Math.abs(row[cur]-row[i]) == cur-i){
                return false;
            }
        }
        return true;
    }
}

AC Python:

class Solution:
    def totalNQueens(self, n: int) -> int:
        self.res = 0
        self.dfs(n, [0] * n, 0)
        return self.res
    
    def dfs(self, n: int, board: List[int], cur: int) -> None:
        if cur == n:
            self.res += 1
            return
        for i in range(n):
            board[cur] = i
            if self.isValid(cur, board):
                self.dfs(n, board, cur + 1)
                
    def isValid(self, cur: int, board: List[int]) -> bool:
        for i in range(0, cur):
            if board[cur] == board[i] or abs(board[cur] - board[i]) == cur - i:
                return False
        
        return True
        

类似N-Queens.