用C语言实现递归和回溯求解数独字段

#include <stdio.h>

#define SIZE 9

// 打印数独的当前状态
void printSudoku(int sudoku[SIZE][SIZE]) {
    for (int i = 0; i < SIZE; i++) {
        for (int j = 0; j < SIZE; j++) {
            printf("%d ", sudoku[i][j]);
        }
        printf("\n");
    }
}

// 检查当前数字是否可以填入指定位置
int isValid(int sudoku[SIZE][SIZE], int row, int col, int num) {
    // 检查行和列是否已经存在该数字
    for (int i = 0; i < SIZE; i++) {
        if (sudoku[row][i] == num || sudoku[i][col] == num) {
            return 0;
        }
    }

    // 检查当前3x3的小方格是否已经存在该数字
    int startRow = row - row % 3;
    int startCol = col - col % 3;
    for (int i = 0; i < 3; i++) {
        for (int j = 0; j < 3; j++) {
            if (sudoku[startRow + i][startCol + j] == num) {
                return 0;
            }
        }
    }

    return 1;
}

// 递归求解数独
int solveSudoku(int sudoku[SIZE][SIZE], int row, int col) {
    // 如果已经遍历完所有行，则表示数独已解
    if (row == SIZE - 1 && col == SIZE) {
        return 1;
    }

    // 如果已经遍历完当前行，则继续下一行的遍历
    if (col == SIZE) {
        row++;
        col = 0;
    }

    // 如果当前位置已经填有数字，则跳过
    if (sudoku[row][col] != 0) {
        return solveSudoku(sudoku, row, col + 1);
    }

    // 尝试填入数字并进行回溯
    for (int num = 1; num <= SIZE; num++) {
        if (isValid(sudoku, row, col, num)) {
            sudoku[row][col] = num;
            if (solveSudoku(sudoku, row, col + 1)) {
                return 1;
            }
            sudoku[row][col] = 0;
        }
    }

    return 0;
}

int main() {
    int sudoku[SIZE][SIZE] = {
        {5, 3, 0, 0, 7, 0, 0, 0, 0},
        {6, 0, 0, 1, 9, 5, 0, 0, 0},
        {0, 9, 8, 0, 0, 0, 0, 6, 0},
        {8, 0, 0, 0, 6, 0, 0, 0, 3},
        {4, 0, 0, 8, 0, 3, 0, 0, 1},
        {7, 0, 0, 0, 2, 0, 0, 0, 6},
        {0, 6, 0, 0, 0, 0, 2, 8, 0},
        {0, 0, 0, 4, 1, 9, 0, 0, 5},
        {0, 0, 0, 0, 8, 0, 0, 7, 9}
    };

    printf("数独的初始状态：\n");
    printSudoku(sudoku);
    printf("\n");

    if (solveSudoku(sudoku, 0, 0)) {
        printf("数独的解：\n");
        printSudoku(sudoku);
    } else {
        printf("无法解决数独。\n");
    }

    return 0;
}