解数独算法，用C语言递归实现

前几天在看离散数学及其应用第七版的时候，看到了数独的相关章节，就自己写了一个数独的解法，先记下来。生成数独的方法以后再写

#include <stdio.h>// 寻找下一个未填充的单元int find_next_empty(int arr[9][9], int startrow, int *row, int *col){int i, j;for (i = startrow; i < 9; i++) for (j = 0; j < 9; j++) if (arr[i][j] == 0) {*row = i;*col = j;return 1;}return 0;}void print(int a[9][9]){int i, j;for (i = 0; i < 9; i++) {for (j = 0; j < 9; j++) {printf("%2d", a[i][j]);if (j == 8) printf("\n");}}}int do_resolve(int arr[9][9], int row, int col){int i, j, n;int next_row, next_col;n = 0;while(1) {next_num:++n;if (n >= 10) break;// 判断行重复for (j = 0; j < 9; j++) {if (arr[row][j] == n) {goto next_num;}}// 判断列重复for (i = 0; i < 9; i++) {if (arr[i][col] == n) {goto next_num;}}/* 判断所在小九宫格重复*/int x = (row / 3) * 3;int y = (col / 3) * 3;for (i = x; i < x + 3; i++) {for (j = y; j < y + 3; j++){if (arr[i][j] == n) {goto next_num;}}}//该单元可以填充arr[row][col] = n;//如果9宫格已填满，完成，这里不考虑有多解的情况if (!find_next_empty(arr, row, &next_row, &next_col)) {return 1;}//否则继续填下一个未填充的格子if (!do_resolve(arr, next_row, next_col)) {arr[row][col] = 0;continue;} else {return 1;}}return 0;}void resolve_sudoku(int a[9][9]){int row, col;find_next_empty(a, 0, &row, &col);do_resolve(a, row, col);}int main(void){int a[9][9] = {8,0,0,0,0,0,0,0,0,0,0,3,6,0,0,0,0,0,0,7,0,0,9,0,2,0,0,0,5,0,0,0,7,0,0,0,0,0,0,0,4,5,7,0,0,0,0,0,1,0,0,0,3,0,0,0,1,0,0,0,0,6,8,0,0,8,5,0,0,0,1,0,0,9,0,0,0,0,4,0,0};resolve_sudoku(a);print(a);return 0;}