力扣第三十七题—

力扣第三十七题——解数独

news2024/9/17 4:03:37

内容介绍

编写一个程序，通过填充空格来解决数独问题。

数独的解法需 遵循如下规则：

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
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）

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1：
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：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
提示：

board.length == 9
board[i].length == 9
board[i][j] 是一位数字或者 '.'
题目数据保证输入数独仅有一个解

完整代码

 bool line[9][9];
bool column[9][9];
bool block[3][3][9];
bool valid;
int* spaces[81];
int spacesSize;

void dfs(char** board, int pos) {
    if (pos == spacesSize) {
        valid = true;
        return;
    }

    int i = spaces[pos][0], j = spaces[pos][1];
    for (int digit = 0; digit < 9 && !valid; ++digit) {
        if (!line[i][digit] && !column[j][digit] && !block[i / 3][j / 3][digit]) {
            line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
            board[i][j] = digit + '0' + 1;
            dfs(board, pos + 1);
            line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = false;
        }
    }
}

void solveSudoku(char** board, int boardSize, int* boardColSize) {
    memset(line, false, sizeof(line));
    memset(column, false, sizeof(column));
    memset(block, false, sizeof(block));
    valid = false;
    spacesSize = 0;

    for (int i = 0; i < 9; ++i) {
        for (int j = 0; j < 9; ++j) {
            if (board[i][j] == '.') {
                spaces[spacesSize] = malloc(sizeof(int) * 2);
                spaces[spacesSize][0] = i;
                spaces[spacesSize++][1] = j;
            } else {
                int digit = board[i][j] - '0' - 1;
                line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
            }
        }
    }

    dfs(board, 0);
}

思路详解

方法概述

solveSudoku：主方法，初始化数据结构并调用DFS方法来填充数独板。
dfs：递归方法，用于尝试填充数独板的每个空格。

数据结构

line：一个二维布尔数组，用于标记每行中每个数字是否已经被使用。
column：一个二维布尔数组，用于标记每列中每个数字是否已经被使用。
block：一个三维布尔数组，用于标记每个3x3宫格中每个数字是否已经被使用。
valid：一个布尔变量，用于标记是否已经找到有效的数独解决方案。
spaces：一个指针数组，用于存储所有空格的位置。
spacesSize：一个整数，用于记录空格的数量。

解题步骤

初始化

使用memset函数将line、column和block数组初始化为false，表示所有数字都未被使用。
将valid变量设置为false，表示尚未找到解决方案。
遍历数独板，对于每个元素：
- 如果是空格（即’.'），则记录该空格的位置到spaces数组，并增加spacesSize。
- 如果是非空格，则更新line、column和block数组，标记该数字已被使用。

深度优先搜索（DFS）

递归方法dfs接收当前填充空格的位置pos。
如果pos等于spacesSize，说明所有空格都已填充，设置valid为true并返回。
对于当前空格位置，尝试填充数字1到9：
- 检查当前数字是否在当前行、当前列和当前3x3宫格中未被使用。
- 如果未被使用，则更新line、column和block数组，将当前数字标记为已使用，并在数独板上填充该数字。
- 递归调用dfs，尝试填充下一个空格。
- 如果递归调用后未找到解决方案，则回溯，撤销当前数字的填充，并继续尝试下一个数字。