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
cpp 复制代码
class Solution {
public:
    bool isvaild(int row,int col,char val,vector<vector<char>>& board){
        //row
        for(int i = 0;i < 9;i++){
            if(board[row][i] == val) return false;
        }
        //col
        for(int j = 0;j < 9;j++){
            if(board[j][col] == val) return false;
        }
        //九宫格
        int startx = (row/3)*3; // 假如在第一个九宫格,row/3=0,再*3=0;
        int starty = (col/3)*3; //假如在第二个九宫格,row/3=1,再*3=3; 我直呼nb
        for(int i = startx;i < startx+3;i++){
            for(int j = starty;j < starty+3;j++){
                if(board[i][j] == val) return false;
            }
        }
        return true;
    }
    bool backtracking(vector<vector<char>>& board){
        for(int i = 0;i < board.size();i++){
            for(int j = 0;j < board[0].size();j++){
                //遇到空格
                if(board[i][j] == '.'){
                    for(char a = '1';a <= '9';a++){
                        //判断这里应该填入啥数字合法
                        if(isvaild(i,j,a,board)){
                            board[i][j] = a;
                            //得将这个状态一直返回
                            if(backtracking(board) == true) return true;
                            board[i][j] = '.'; // 回溯
                        }
                    }
                    return false; //填入0-9都不对,都不合法,填错了。
                }
            }
        }
        return true; //填完且填正确了。
    }
    void solveSudoku(vector<vector<char>>& board) {
        backtracking(board);
    }
};
相关推荐
牧羊人.3331 小时前
卷积神经网络之VGG
人工智能·深度学习·神经网络·算法·cnn
akinggw1 小时前
libsherpacore - 基于 sherpa-onnx 的 C++ 语音交互静态库
c++·交互·语音识别
xiaoye-duck1 小时前
《C++ 项目:手写大模型接入 SDK》ChatGPT接口设计(中):全量响应实现和测试
网络·c++·语言模型·chatgpt
小郎君6581 小时前
Vue 多标签页的加载态设计:为什么我没有用 Suspense 实现 PageLoading
vue.js·算法
JJJennie7772 小时前
MAI Gateway私有化部署全攻略:大模型网关如何接入?附架构图
算法·gateway·ai网关·魔芋ai
東隅已逝,桑榆非晚2 小时前
算法3.双指针法-快乐数
c++·笔记·学习·算法
专注仿真2 小时前
卷积神经网络(CNN)算法详解
人工智能·python·算法·代码·大模型训练
FII工业富联科技服务2 小时前
多智能体如何进入制造现场?从业务协同到数字孪生验证
大数据·人工智能·算法·机器人·制造·数字孪生
≮傷£≯√2 小时前
httpserver_QNetWorkClient
算法
Evand J3 小时前
【PMSM例程11】永磁同步电机(PMSM)基于UKF的转速与转子位置估计。附下载链接
算法·matlab·卡尔曼滤波·永磁同步电机·ukf·pmsm·位置估计