题目:

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

![这里写图片描述](http://img.blog.csdn.net/20160409183641502)
A partially filled sudoku which is valid.

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

思路:

  • 首先这是一道数独题目,关于数独的性质,可以参考下面的链接

    数独规则

  • 考虑根据数独的各个条件来逐个解决,需要判断每行,每一列,以及每个小方格是否是1~9,出现一次,思路是建立一个list添加,判断是否出现重复,出现返回false;同样如果board == null,或者行数和列数不等于9,也返回false

  • -

代码:

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        ArrayList<ArrayList<Character>> rows = new ArrayList<ArrayList<Character>>();
        ArrayList<ArrayList<Character>> cols = new ArrayList<ArrayList<Character>>();
        ArrayList<ArrayList<Character>> boxs = new ArrayList<ArrayList<Character>>();
        if(board == null || board.length != 9|| board[0].length != 9){
            return false;
        }
        for(int i = 0;i < 9;i++){
            rows.add(new ArrayList<Character>());
            cols.add(new ArrayList<Character>());
            boxs.add(new ArrayList<Character>());
        }
        for(int a = 0;a < board.length;a++){
            for(int b = 0;b < board[0].length;b++){
                if(board[a][b] == '.'){
                    continue;
                }
                ArrayList<Character> row = rows.get(a);
                if(row.contains(board[a][b])){
                    return false;
                }else{
                    row.add(board[a][b]);
                }
                ArrayList<Character> col = cols.get(b);
                if(col.contains(board[a][b])){
                    return false;
                }else{
                    col.add(board[a][b]);
                }
                ArrayList<Character> box = boxs.get(getNum(a,b));
                if(box.contains(board[a][b])){
                    return false;
                }else{
                    box.add(board[a][b]);
                }
            }
        }return true;
    }
    public int getNum(int i,int j){
        return (i/3)*3+j/3;
    }
}

LeetCode(38)-Valid Sudoku的更多相关文章

  1. [LeetCode] 036. Valid Sudoku (Easy) (C++)

    指数:[LeetCode] Leetcode 解决问题的指数 (C++/Java/Python/Sql) Github: https://github.com/illuz/leetcode 036. ...

  2. LeetCode:36. Valid Sudoku,数独是否有效

    LeetCode:36. Valid Sudoku,数独是否有效 : 题目: LeetCode:36. Valid Sudoku 描述: Determine if a Sudoku is valid, ...

  3. LeetCode 36 Valid Sudoku

    Problem: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board ...

  4. 【leetcode】Valid Sudoku

    题目简述: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board cou ...

  5. Java [leetcode 36]Valid Sudoku

    题目描述: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board cou ...

  6. [leetcode] 20. Valid Sudoku

    这道题目被放在的简单的类别里是有原因的,题目如下: Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. ...

  7. 蜗牛慢慢爬 LeetCode 36.Valid Sudoku [Difficulty: Medium]

    题目 Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku board could ...

  8. leetCode 36.Valid Sudoku(有效的数独) 解题思路和方法

    Valid Sudoku Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudoku bo ...

  9. LeetCode 036 Valid Sudoku

    题目要求:Valid Sudoku Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules. The Sudo ...

随机推荐

  1. 剑指offer面试题5 从头到尾打印链表(java)

    注:(1)这里体现了java数据结构与C语言的不同之处 (2)栈的操作直接利用stack进行 package com.xsf.SordForOffer; import java.util.Stack; ...

  2. socket系列之服务器端socket——ServerSocket类

    一般地,Socket可分为TCP套接字和UDP套接字,再进一步,还可以被分为服务器端套接字跟客户端套接字.这节我们先关注TCP套接字的服务器端socket,Java中ServerSocket类与之相对 ...

  3. Hessian源码分析--HessianServlet

    Hessian可以通过Servlet来对外暴露服务,HessianServlet继承于HttpServlet,但这仅仅是一个外壳,使用web服务器来提供对外的Http请求,在web.xml中我们会进行 ...

  4. 成员函数的const到底修饰的是谁

    demo <pre name="code" class="cpp">class Test { public: const void OpVar(in ...

  5. A*寻路算法入门(二)

    大熊猫猪·侯佩原创或翻译作品.欢迎转载,转载请注明出处. 如果觉得写的不好请告诉我,如果觉得不错请多多支持点赞.谢谢! hopy ;) 免责申明:本博客提供的所有翻译文章原稿均来自互联网,仅供学习交流 ...

  6. 怎么在Eclipse中添加VI插件

    下载地址 Vi插件下载位置 怎么安装? 将下载下来的zip文件进行解压,然后把对于的目录下的文件分别复制到eclipse目录下的plugins 和features目录下: 注册 在eclipse根目录 ...

  7. Android官方命令深入分析之hprof-conv

    hprof-conv工具可以将Android SDK工具生成的HPROF文件生成一个标准的格式,这样你就可以使用工具进行查看: hprof-conv [-z] <infile> <o ...

  8. Linux多线程实践(6) --Posix读写锁解决读者写者问题

    Posix读写锁 int pthread_rwlock_init(pthread_rwlock_t *restrict rwlock, const pthread_rwlockattr_t *rest ...

  9. 学习tornado:异步

    why asynchronous tornado是一个异步web framework,说是异步,是因为tornado server与client的网络交互是异步的,底层基于io event loop. ...

  10. javascript语法之Date对象与小案例

    <!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/ ...