leetcode：N-Queens II 【Java】-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解决方案配置，还计算了不同解决方案的数量。通过使用回溯算法，文章提供了一个Java实现示例，展示了如何有效地确定任意大小棋盘上放置皇后的所有非冲突布局。

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一、问题描述

Follow up for N-Queens problem.

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

二、问题分析

参考问题leetcode：N-Queens 【Java】

三、算法代码

public class Solution {
    public int totalNQueens(int n) {
        int [] total = new int [] {0};
        if (n <= 0) return total[0];
        char[][] board = new char[n][n];
        for (char[] row : board) {
            Arrays.fill(row, '.');
        }
        boolean[] col_occupied = new boolean[n];
        placeQueen(total, board, col_occupied, 0, n);
        return total[0];
    }
    private void placeQueen(int [] total, char[][] board, boolean[] col_occupied, int rowNum, int n) {
        if (rowNum == n) {
            total[0]++;
            return;
        }
        for (int colNum = 0; colNum < n; colNum++) {
            if (isValid(board, col_occupied, rowNum, colNum, n)){
                board[rowNum][colNum] = 'Q';
                col_occupied[colNum] = true;
                placeQueen(total, board, col_occupied, rowNum + 1, n);
                board[rowNum][colNum] = '.'; //回溯，尝试皇后rowNum的下一个位置
                col_occupied[colNum] = false;
            }
        }
    }
    private boolean isValid(char[][]board, boolean[] col_occupied, int row, int col, int n) {
        if (col_occupied[col]) return false;
        for (int i = 1; row - i >= 0 && col - i >= 0; i++) {//反对角斜线
            if (board[row - i][col - i] == 'Q') return false;
        }
        for (int i = 1; row - i >= 0 && col + i < n; i++) {//正对角斜线
            if (board[row - i][col + i] == 'Q') return false;
        }
        return true;
    }
}