学渣带你刷Leetcode0036有效的数独

带你刷Leetcode0036有效的数独"/>

学渣带你刷Leetcode0036有效的数独

题目描述

判断一个 9x9 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。

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。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1:

输入:
[
  ["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
示例 2:

输入:
[
  ["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。
说明:

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。
给定数独序列只包含数字 1-9 和字符 '.' 。
给定数独永远是 9x9 形式的。

来源：力扣（LeetCode）
链接：
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白话题目：

[suˈdəʊkuː],判断已经填入的状态是否有效。就是同行同列，同九宫格不能有重复的1，2，3,4,5,6,7,8,9。

算法：

（1）输入测试，就是从一个txt文档中读入输入  用到了fopen，fgets，strtok（用空格分隔字符串）

（2）存入char **board中，注意不能用board[][],当初写的时候出现了问题

（3）判断

//1，循环判断行是否都有效
//2，循环判断列是否都有效
//3，循环判断每个9宫格是否都有效
//三者都有效时整个数独才有效

思路简单，容易理解，完成的就是代码实现的事，至于后面的37题，我们再想办法。

今朝有酒今朝醉，明日愁来明日愁。

详细解释关注 B站  【C语言全代码】学渣带你刷Leetcode 不走丢 

C语言完全代码

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#define BUF_SIZE (200)
bool isValidLine(char** board, int line)
{int iNumFlag[9]= {0};int index= 0;bool bRet= true;for (index = 0; index < 9; index++){if ((board[line][index] >= '1') && (board[line][index] <= '9')){if (1 == iNumFlag[board[line][index] - '1']){bRet = false;return bRet;}else{iNumFlag[board[line][index] - '1'] = 1;}}}return bRet;
}bool isValidColumn(char** board, int column)
{int     iNumFlag[9]     = {0};int     index           = 0;bool    bRet            = true;for (index = 0; index < 9; index++){if ((board[index][column] >= '1') && (board[index][column] <= '9')){if (1 == iNumFlag[board[index][column] - '1']){bRet = false;return bRet;}else{iNumFlag[board[index][column] - '1'] = 1;}}}return bRet;
}bool isvalidSpace(char** board, int line, int column)
{int     iNumFlag[9]     = {0};int     i               = 0;int     j               = 0;bool    bRet            = true;for (i = 0; i < 3; i++){for (j = 0; j < 3; j++){if ((board[line + i][column + j] >= '1') && (board[line + i][column + j] <= '9')){if (1 == iNumFlag[board[line + i][column + j] - '1']){bRet = false;return bRet;}else{iNumFlag[board[line + i][column + j] - '1'] = 1;}}}}return bRet;
}bool isValidSudoku(char** board, int boardSize, int* boardColSize)
{int     i       = 0;int     j       = 0;bool    bRet    = true;printf("start\n");//行判断for (i = 0; i < 9; i++){if (false == isValidLine(board, i)){bRet = false;return bRet;}}//列判断for (j = 0; j < 9; j++){if (false == isValidColumn(board, j)){bRet = false;return bRet;}}//9宫格判断for (i = 0; i < 9; i += 3){for (j = 0; j < 9; j += 3){if (false == isvalidSpace(board, i, j)){bRet = false;return bRet;}}}return bRet;
}int main()
{int boardSize=9;int boardColSize=9;char** board=(char**)malloc(sizeof(char*)*boardSize);char arr[9][BUF_SIZE]= {0};FILE *fp=fopen("suduku.txt","r");if(!fp){printf("open fail\n");}int i=0;int j=0;char buf[BUF_SIZE]= {'0'};while((fgets(arr[i],BUF_SIZE,fp))!=NULL){puts(arr[i]);   //每一行都被存在了arr[i]中board[i]=(char*)malloc(sizeof(char)*boardColSize);char *subarr= strtok(arr[i]," ");int  j=0;while(subarr!=NULL){// printf( "---%d\n", strlen(subarr));//  printf( "---%s\n", subarr );board[i][j]=subarr[0];  //拆分的是字符串，  回车也有，我们就取一个第一位就好了subarr =strtok(NULL," ");//继续j++;}i++;}printf("\n");fclose(fp);for(i=0; i<boardSize; i++){for(j=0; j<boardColSize; j++){printf("%c",board[i][j]);}printf("\n");}bool result=isValidSudoku(board, boardSize,&boardColSize);printf("%d\n", result);return 0;
}