「力扣」第 36 题:有效的数独(中等)

liweiwei1419 ... 2021-12-27 哈希表
  • 哈希表
About 3 min

# 题目描述

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ,验证已经填入的数字是否有效即可。

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。(请参考示例图)

注意:

  • 一个有效的数独(部分已被填充)不一定是可解的。
  • 只需要根据以上规则,验证已经填入的数字是否有效即可。
  • 空白格用 '.' 表示。

示例 1:

img

输入:board =
[["5","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
输出:true
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示例 2:

输入:board =
[["8","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
输出:false
解释:除了第一行的第一个数字从 5 改为 8 以外,空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
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提示:

  • board.length == 9
  • board[i].length == 9
  • board[i][j] 是一位数字(1-9)或者 '.'

# 思路分析

在遍历的时候,只要遇到的是数字,就给对应的行、列、box 加上标记,表示已经占位。

image.png

图片来自 官方题解 (opens new window)

# 方法:布尔数组(哈希表)

参考代码 1:box 二维表格,重点理解 int boardIndex = (i / 3) * 3 + j / 3; 这行代码。

public class Solution {

    // 知识点:哈希表,空间换时间

    public boolean isValidSudoku(char[][] board) {
        // 设置成为 10 是为了照顾到数字 9 的情况(下标 9 数字需要到 10)
        // 第 1 维表示行的下标
        boolean[][] row = new boolean[9][10];
        // 第 1 维表示列的下标
        boolean[][] col = new boolean[9][10];
        // 第 1 维表示 board 的下标
        boolean[][] box = new boolean[9][10];

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                // 只验证数字,因此 . 跳过
                if (board[i][j] == '.') {
                    continue;
                }

                // 提取出数字
                int num = board[i][j] - '0';
                // 重点:计算在第几格
                int boardIndex =  (i / 3) * 3 + j / 3;

                // 如果发现冲突,直接返回 false
                if (row[i][num] || col[j][num] || box[boardIndex][num]) {
                    return false;
                }

                // 然后占住位置
                row[i][num] = true;
                col[j][num] = true;
                box[boardIndex][num] = true;
            }
        }
        return true;
    }
}
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参考代码 2

public class Solution {

    // 知识点:哈希表,空间换时间

    public boolean isValidSudoku(char[][] board) {
        // 设置成为 10 是为了照顾到数字 9 的情况(下标 9 数字需要到 10)
        // 第 1 维表示行的下标
        boolean[][] row = new boolean[9][10];
        // 第 1 维表示列的下标
        boolean[][] col = new boolean[9][10];
        // 第 1 维表示 board 的下标
        boolean[][][] box = new boolean[3][3][10];

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                // 只验证数字,因此 . 跳过
                if (board[i][j] == '.') {
                    continue;
                }

                // 提取出数字
                int num = board[i][j] - '0';
                // 重点:计算在第几格
                int boardIndex = (i / 3) * 3 + j / 3;

                // 如果发现冲突,直接返回 false
                if (row[i][num] || col[j][num] || box[i / 3][j / 3][num]) {
                    return false;
                }

                // 然后占住位置
                row[i][num] = true;
                col[j][num] = true;
                box[i / 3][j / 3][num] = true;
            }
        }
        return true;
    }
}
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扩展,可以考虑使用位运算的相关技巧,代替布尔数组。

Last update: January 14, 2022 00:02
Contributors: liweiwei1419