389. Valid Sudoku【LintCode java】
Description
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.
Clarification
Example
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】的更多相关文章
- 423. Valid Parentheses【LintCode java】
Description Given a string containing just the characters '(', ')', '{', '}', '[' and ']', determine ...
- 415. Valid Palindrome【LintCode java】
Description Given a string, determine if it is a palindrome, considering only alphanumeric character ...
- 376. Binary Tree Path Sum【LintCode java】
Description Given a binary tree, find all paths that sum of the nodes in the path equals to a given ...
- 372. Delete Node in a Linked List【LintCode java】
Description Implement an algorithm to delete a node in the middle of a singly linked list, given onl ...
- 451. Swap Nodes in Pairs【LintCode java】
Description Given a linked list, swap every two adjacent nodes and return its head. Example Given 1- ...
- 445. Cosine Similarity【LintCode java】
Description Cosine similarity is a measure of similarity between two vectors of an inner product spa ...
- 433. Number of Islands【LintCode java】
Description Given a boolean 2D matrix, 0 is represented as the sea, 1 is represented as the island. ...
- 422. Length of Last Word【LintCode java】
Description Given a string s consists of upper/lower-case alphabets and empty space characters ' ', ...
- 420. Count and Say【LintCode java】
Description The count-and-say sequence is the sequence of integers beginning as follows: 1, 11, 21, ...
随机推荐
- oracle系列(四)PL/SQL
过程,函数,触发器是PL/SQL编写的,存储在oracle中的.PL/SQL是非常强大的数据库过程语言. PL/SQL优点:性能,模块化,网络传输量,安全性缺点:移植性不好 简单分类:块:过程,函数, ...
- 选择排序_c++
选择排序_c++ GitHub 文解 选择排序的核心思想是对于 N 个元素进行排序时,对其进行 K = (N - 1) 次排序,每次排序从后(N + 1 - K)个数值中选择最小的元素与以 (K - ...
- DB数据源之SpringBoot+MyBatis踏坑过程(四)没有使用连接池的后果
DB数据源之SpringBoot+MyBatis踏坑过程(四)没有使用连接池的后果 liuyuhang原创,未经允许禁止转载 系列目录连接 DB数据源之SpringBoot+Mybatis踏坑过程实 ...
- 811. Subdomain Visit Count (5月23日)
解答 class Solution { public: vector<string> subdomainVisits(vector<string>& cpdomains ...
- Django学习笔记2
1.BookInfo.objects.all() objects:是Manager类型的对象,用于与数据库进行交互 当定义模型类时没有指定管理器,则Django会为模型类提供一个名为objects的管 ...
- MapReduce清洗日志数据统计PV量
package mapreduce.webpv; import java.io.IOException; import org.apache.commons.lang.StringUtils; imp ...
- 基于CentOS-6.9_x64系统QT环境搭建
想从事QT开发的人员,首先要做的第一件事就是开发环境的搭建.本人也是一位刚入门的新手,为了搭建这么一个环境,参考了很多的网上教程,然而中间依然走了不少弯路.现将过程记录下来. 一.开发环境 Ce ...
- Waltz of love
Waltz of love Love me tenderly Love me softly Close your eyes,fling to the dangcing hall Follow your ...
- Python值集合操作
list_l1 = [12,1,4,6,9,5,4,2,11] list_l1 = set(list_l1) print(list_l1,type(list_l1)) #集合也是无序的 list_2 ...
- 解决区间第K大的问题的各种方法
例题:http://poj.org/problem?id=2104 最近可能是念念不忘,必有回响吧,总是看到区间第k大的问题,第一次看到是在知乎上有人面试被弄懵了后来又多次在比赛中看到.以前大概是知道 ...