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。
1-9
在每一行只能出现一次。1-9
在每一列只能出现一次。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:
1-9
和字符 '.'
。9x9
形式的。解法一:非常巧妙!!!利用三个二维数组
class Solution {
public:
bool isValidSudoku(vector>& board) {
int row[9][9] = {0}, col[9][9] = {0}, box[9][9] = {0};
for (int i = 0; i != board.size(); ++i){
for (int j = 0; j != board[i].size(); ++j){
if (board[i][j] != '.')
{
int num = board[i][j] - '0' - 1, k = i / 3 * 3 + j / 3;
if (row[i][num] || col[j][num] || box[k][num])
return false;
row[i][num] = col[j][num] = box[k][num] = 1;
}
}
}
return true;
}
};
解法二:利用自定义函数,很繁琐。。。
class Solution {
public:
bool isvalid(vector vec){//判断9个元素中是否有重复的
for(int i = 0;i < vec.size()-1;++i){
for(int j = i+1;j < vec.size();++j){
if(vec[i]==vec[j] && vec[i]!='.') return false;
}
}
return true;
}
vector fun(int a,int b,vector> board){//将方格型的9个元素写到vector中
vector v(9);
int k=0;
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
v[k++]=board[a+i][b+j];
}
}
return v;
}
bool isValidSudoku(vector>& board) {
vector v(9);
for(int i=0;i<9;++i){//将竖着的9个元素写成vector,添加到board里
for(int j=0;j<9;++j){
v[j]=board[j][i];
}
board.push_back(v);
}
for(int i=0;i<7;i+=3){//将9个9宫格加入到board里
for(int j=0;j<7;j+=3){
board.push_back(fun(i,j,board));
}
}
for(int n=0;n