数独 深度优先 递归实现 简单明了

#include <stdio.h>
#include <stdlib.h>
char s[9][9]= 
{{'8','.','.','.','.','.','.','.','.'},{'.','.','3','6','.','.','.','.','.'},
{'.','7','.','.','9','.','2','.','.'},{'.','5','.','.','.','7','.','.','.'},
{'.','.','.','.','4','5','7','.','.'},{'.','.','.','1','.','.','.','3','.'},
{'.','.','1','.','.','.','.','6','8'},{'.','.','8','5','.','.','.','1','.'},
{'.','9','.','.','.','.','4','.','.'}};
int* getExistedNums(char board[][9],int i,int j){
    int k;
    int* ExistedNums = (int*)calloc(10,sizeof(int));
    for(k=0;k<9;k++){
        if(board[i][k]!='.'&&!ExistedNums[board[i][k]-'0']) ExistedNums[board[i][k]-'0']=1;
		if(board[k][j]!='.'&&!ExistedNums[board[k][j]-'0']) ExistedNums[board[k][j]-'0']=1;
    }
    for(k=3*(i/3);k<3*(i/3)+3;k++){
        for(int l=3*(j/3);l<3*(j/3)+3;l++){
            if(board[k][l]!='.'&&!ExistedNums[board[k][l]-'0']) ExistedNums[board[k][l]-'0']=1;
        }
    }
    return ExistedNums;
}
bool SudoKu(char board[][9],int i,int j){
    if(i==9)return true;
    if(board[i][j]!='.'){
        if(SudoKu(board,(j==8?i+1:i),(j==8?0:j+1)))return true;
    }else{
        int* ExistedNums=getExistedNums(board,i,j);
        for(int k=1;k<10;k++){
            if(ExistedNums[k]==0){
                board[i][j]=k+48;
                if(SudoKu(board,(j==8?i+1:i),(j==8?0:j+1)))return true;
                board[i][j]='.';//trace back
            }
        }
        free(ExistedNums);
    }
    return false;
} 
int main(int argc, char* argv[]) {
	SudoKu(s,0,0);
	for (int i=0; i<9; i++) {
		for (int j=0; j<9; j++) {
			printf("%c\t",s[i][j]);
		}
		printf("\n");
	}
	return 0;
}

可以借鉴这个哥们的check进行剪枝，剪枝的有时间再写