37. 解数独

编写一个程序,通过填充空格来解决数独问题。

数独的解法需遵循如下规则

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 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)

数独部分空格内已填入了数字,空白格用 '.' 表示。

示例 1:

复制代码
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:[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释:输入的数独如上图所示,唯一有效的解决方案如下所示:

提示:

  • board.length == 9
  • board[i].length == 9
  • board[i][j] 是一位数字或者 '.'
  • 题目数据 保证 输入数独仅有一个解
java 复制代码
    public boolean solveSudoku(char[][] board) {
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                如果为空格就依次把1-9填入
                if (board[i][j] == '.') {
                    for (char k = '1'; k <= '9'; k++) {
                        // 判断填入的数字是否合法
                        if (isValid(i, j, k, board)) {
                            board[i][j] = k;
                            // 递归
                            boolean result = solveSudoku(board);
                            if(result == true)  
                                return true;
                            board[i][j]='.';  //不合法就回溯,回溯前先重新赋值为空
                        }
                    }
                    // 填写完成后返回
                    return false;
                }
            }
        }
        return true;
    }

    /**
     * 判断放入棋盘的数字是否合法
     */
    public boolean isValid(int row, int col, char val, char[][] board) {
        // 验证一行
        for (int i = 0; i < 9; i++) { // 判断行里是否重复
            if (board[row][i] == val) {
                return false;
            }
        }

        // 验证一列
        for (int j = 0; j < 9; j++) { // 判断行里是否重复
            if (board[j][col] == val) {
                return false;
            }
        }

        //    验证3*3
        int startRow = (row / 3) * 3;
        int startCol = (col / 3) * 3;
        for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复
            for (int j = startCol; j < startCol + 3; j++) {
                if (board[i][j] == val ) {
                    return false;
                }
            }
        }
        return true;
    }
相关推荐
郝学胜-神的一滴2 分钟前
Effective Python 条款 10 :海象运算符_=
开发语言·python·程序人生·开源
wuyk5554 分钟前
Python零基础入门第十四章:异常处理(try-except)
开发语言·python
wuyk5559 分钟前
【Socket 进阶之路】第 3 章 TCP 三次握手 & 四次挥手深度剖析|连接建立、断开、状态机、TIME_WAIT 核心工程问题
服务器·开发语言·网络·物联网·网络协议·tcp/ip
彧azz6 小时前
图的存储结构详解:邻接矩阵的原理、实现与应用
开发语言·数据结构·学习·php
嵌入式学习菌6 小时前
Modbus‑RTU 数据类型分析
开发语言·单片机·bug
matlab代码7 小时前
基于matlab数字图像处理的指纹识别系统【源码68期】
开发语言·matlab
matlab代码7 小时前
基于matlab的水果识别系统(苹果香蕉菠萝梨桃子)【源码65期】
开发语言·matlab
wuyk5558 小时前
从零吃透 MQTT 通信|第 8 章 FreeRTOS 多任务架构下 MQTT 工程架构,任务拆分、队列解耦、临界区保护
c语言·开发语言·stm32·学习·架构
weixin199701080168 小时前
[特殊字符]《跨境二手ERP对接Back Market:标准化API + 7~10工作日技术合规审核实录》(附Python源码)
开发语言·python·pandas
尘客-追梦8 小时前
Day 01:插件化之前,先看清 ABI 这堵墙
开发语言·c++·qt