标签：lintcode file character empty HERE param follow font png
Determine whether a Sudoku is valid.
The Sudoku board could be partially filled, where empty cells are filled with the character ..
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
The following partially filed sudoku is valid.
解题：判断数独是否有效。如果每一行、每一列、每一个小方块里的数不重复，就算有效。当然，前提必须是1~9之间。此题空白部分是用“ . ”表示的，要注意一下。具体做法为，检查行、列、方块，把这些数拿出来，放在一个一维数组中，判断这个一维数组里的数，满不满足相应的条件。代码如下：
public class Solution {
/**
* @param board: the board
* @return: whether the Sudoku is valid
*/
public boolean isValidSudoku(char[][] board) {
// write your code here
char[]temp = new char[9];
for(int i = 0; i < 9; i++){
//检查行
for(int j = 0; j < 9; j++){
temp[j] = board[i][j];
}
if(judge(temp) == false){
return false;
}
//检查列
for(int j = 0; j < 9; j++){
temp[j] = board[j][i];
}
if(judge(temp) == false){
return false;
}
}
//小格子
for(int i = 0; i < 3; i++){
for(int j = 0; j < 3; j++){
int b = 0;
for(int k = 0 + i * 3; k < 3 + i * 3; k++){
for(int m = 0 + j * 3; m < 3 + j * 3; m++){
temp[b++] = board[k][m];
}
}
if(judge(temp) == false)
return false;
}
}
return true;
}
public boolean judge(char[]temp){
for(int i =0; i < 9; i++){
if(temp[i] > ‘9‘ || temp[i] < ‘1‘ ){
if(temp[i] != ‘.‘)
return false;
}
}
for(int i = 1; i < 9; i++){
for(int j = 0; j < i; j++){
if(temp[i] == temp[j] ){
if(temp[i] != ‘.‘)
return false;
}
}
}
return true;
}
}
389. Valid Sudoku【LintCode java】
标签：lintcode file character empty HERE param follow font png
原文地址：https://www.cnblogs.com/phdeblog/p/9313231.html
