请你判断一个 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。(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。
注意:
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。
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:true
示例 2:
输入:board =
[["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。
#include<iostream> #include<stack> #include<algorithm> #include<string> #include<vector> using namespace std; /* 由于board中的整数限定在1到9的范围内,因此可以分别建立哈希表来存储任一个数在相应维度上是否出现过。 维度有3个:所在的行,所在的列,所在的box,注意box的下标也是从左往右、从上往下的。 遍历到每个数的时候,例如boar[i][j],我们判断其是否满足三个条件: 在第 i 个行中是否出现过 在第 j 个列中是否出现过 在第 j/3 + (i/3)*3个box中是否出现过.为什么是j/3 + (i/3)*3呢? */ class Solution { public: bool isValidSudoku(vector<vector<char>>& board) { int row[9][10] = { 0 };// 哈希表存储每一行的每个数是否出现过,默认初始情况下,每一行每一个数都没有出现过 // 整个board有9行,第二维的维数10是为了让下标有9,和数独中的数字9对应。 int col[9][10] = { 0 };// 存储每一列的每个数是否出现过,默认初始情况下,每一列的每一个数都没有出现过 int box[9][10] = { 0 };// 存储每一个box的每个数是否出现过,默认初始情况下,在每个box中,每个数都没有出现过。整个board有9个box。 for (int i = 0; i<9; i++) { for (int j = 0; j<9; j++) { // 遍历到第i行第j列的那个数,我们要判断这个数在其所在的行有没有出现过, // 同时判断这个数在其所在的列有没有出现过 // 同时判断这个数在其所在的box中有没有出现过 if (board[i][j] == '.') { continue; } int curNumber = board[i][j] - '0'; if (row[i][curNumber]) { return false; } if (col[j][curNumber]) { return false; } if (box[j / 3 + (i / 3) * 3][curNumber]) { return false; } row[i][curNumber] = 1;// 之前都没出现过,现在出现了,就给它置为1,下次再遇见就能够直接返回false了。 col[j][curNumber] = 1; box[j / 3 + (i / 3) * 3][curNumber] = 1; } } return true; } };