zoukankan      html  css  js  c++  java
  • LeetCode 36. 有效的数独

    题目描述

    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。

    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。


    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:

    • 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。
    • 给定数独序列只包含数字 1-9 和字符 '.' 。
    • 给定数独永远是 9x9 形式的。

    来源:力扣(LeetCode)
    链接:https://leetcode-cn.com/problems/valid-sudoku

    思路解析

    暴力检查即可。

    代码实现

    class Solution {
    private:
        bool Check(vector<vector<char>>& board) {
            set<char> colNumSet;
            set<char> rowNumSet;
            for(int i = 0; i < board.size(); i++) {
                colNumSet.clear();
                rowNumSet.clear();
                for(int j = 0; j < board.size(); j++) {
                    if(board[j][i] != '.') {
                        if(!colNumSet.insert(board[j][i]).second)
                            return false;
                    }
                    if(board[i][j] != '.') {
                        if(!rowNumSet.insert(board[i][j]).second)
                            return false;
                    }
                }
            }
            return true;
        }
    public:
        bool isValidSudoku(vector<vector<char>>& board) {
            for(int I = 0; I < 3; I++) {
                for(int J = 0; J < 3; J++) {
                    set<char> blockNumSet;
                    for(int i = 0; i < 3; i++) {
                        for(int j = 0; j < 3; j++) {
                            if(board[I * 3 + i][J * 3 + j] != '.') {
                                if(!blockNumSet.insert(board[I * 3 + i][J * 3 + j]).second)
                                    return false;
                            }
                        }
                    }
                }
            }
            return Check(board);
        }
    };
    
  • 相关阅读:
    Java并发基础10:原子性操作类的使用
    Java并发基础09. 多个线程间共享数据问题
    Java并发基础08. 造成HashMap非线程安全的原因
    Java并发基础07. ThreadLocal类以及应用技巧
    Java并发基础06. 线程范围内共享数据
    Java并发基础05. 传统线程同步通信技术
    Java并发基础04. 线程技术之死锁问题
    我是如何从通信转到Java软件开发工程师的?
    IOS 判断耳机插入/拔出
    海量数据处理
  • 原文地址:https://www.cnblogs.com/xqmeng/p/13894491.html
Copyright © 2011-2022 走看看