有效的数独——数组、哈希表、矩阵
目录
该题目来自：https://leetcode-cn.com/problems/valid-sudoku/
题目
36. 有效的数独
难度中等786收藏分享切换为英文接收动态反馈
请你判断一个 9 x 9 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）
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。
- 空白格用
'.'表示。
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。
提示：
board.length == 9board[i].length == 9board[i][j]是一位数字（1-9）或者'.'
题解
呜呜呜，像我这种臭鱼烂虾也配做出leetcode中等难度的题了吗？
我的方法，分别扫描行、扫描列、扫描九宫格，三次遍历得出结果：
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
unordered_set<int> hashset;
// 扫描行
for(int i=0;i<9;i++){
for(int j=0;j<9;j++){
if(board[i][j]!='.'){
if(hashset.find(board[i][j])!=hashset.end()){
// cout << "出问题了" << i << j << endl;
return false;
}
hashset.insert(board[i][j]);
}
}
hashset.clear();
}
// 扫描列
for(int j=0;j<9;j++){
for(int i=0;i<9;i++){
if(board[i][j]!='.'){
if(hashset.find(board[i][j])!=hashset.end()){
// cout << "出问题了" << i << j << endl;
return false;
}
hashset.insert(board[i][j]);
}
}
hashset.clear();
}
// 扫描九宫格
for(int m=0;m<3;m++){
for(int n=0;n<3;n++){ //拆成3*3的九宫格
for(int i=m*3;i<(m+1)*3;i++){
for(int j=n*3;j<(n+1)*3;j++){
if(board[i][j]!='.'){
if(hashset.find(board[i][j])!=hashset.end()){
// cout << "出问题了" << i << j << endl;
return false;
}
hashset.insert(board[i][j]);
}
}
}
hashset.clear();
}
}
return true;
}
};
大佬的方法，一次遍历就行了：
class Solution {
public:
bool isValidSudoku(vector<vector<char>>& board) {
int rows[9][9];
int columns[9][9];
int subboxes[3][3][9];
memset(rows,0,sizeof(rows));
memset(columns,0,sizeof(columns));
memset(subboxes,0,sizeof(subboxes));
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char c = board[i][j];
if (c != '.') {
int index = c - '0' - 1;
rows[i][index]++;
columns[j][index]++;
subboxes[i / 3][j / 3][index]++;
if (rows[i][index] > 1 || columns[j][index] > 1 || subboxes[i / 3][j / 3][index] > 1) {
return false;
}
}
}
}
return true;
}
};
作者：LeetCode-Solution
链接：https://leetcode-cn.com/problems/valid-sudoku/solution/you-xiao-de-shu-du-by-leetcode-solution-50m6/
来源：力扣（LeetCode）
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节省遍历次数的关键是创建了三个数组，
rows[i][index]记录当前数字index在第i行出现的次数。columns[j][index]记录当前数字index在第j列出现的次数。subboxes[i/3][j/3][index]记录当前数字index在第[i/3][j/3]个九宫格中出现的次数。
因为有确定的遍历次数和确定的数组大小，所以时间和空间复杂度都是O(1)