• 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:

    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
    1. class Solution {
    2. public:
    3. bool isvaild(int row,int col,char val,vectorchar>>& board){
    4. //row
    5. for(int i = 0;i < 9;i++){
    6. if(board[row][i] == val) return false;
    7. }
    8. //col
    9. for(int j = 0;j < 9;j++){
    10. if(board[j][col] == val) return false;
    11. }
    12. //九宫格
    13. int startx = (row/3)*3; // 假如在第一个九宫格,row/3=0,再*3=0;
    14. int starty = (col/3)*3; //假如在第二个九宫格,row/3=1,再*3=3; 我直呼nb
    15. for(int i = startx;i < startx+3;i++){
    16. for(int j = starty;j < starty+3;j++){
    17. if(board[i][j] == val) return false;
    18. }
    19. }
    20. return true;
    21. }
    22. bool backtracking(vectorchar>>& board){
    23. for(int i = 0;i < board.size();i++){
    24. for(int j = 0;j < board[0].size();j++){
    25. //遇到空格
    26. if(board[i][j] == '.'){
    27. for(char a = '1';a <= '9';a++){
    28. //判断这里应该填入啥数字合法
    29. if(isvaild(i,j,a,board)){
    30. board[i][j] = a;
    31. //得将这个状态一直返回
    32. if(backtracking(board) == true) return true;
    33. board[i][j] = '.'; // 回溯
    34. }
    35. }
    36. return false; //填入0-9都不对,都不合法,填错了。
    37. }
    38. }
    39. }
    40. return true; //填完且填正确了。
    41. }
    42. void solveSudoku(vectorchar>>& board) {
    43. backtracking(board);
    44. }
    45. };

  • 相关阅读:
    【微服务|Sentinel】SentinelResourceAspect详解
    经济数据预测 | Python实现ELM极限学习机股票价格时间序列预测
    C //例5.11 译密码。为使电文保密,往往按一定规律将其转换成密码,收报人再按约定的规律将其译回原文。
    Lua语法入门
    《羊了个羊》微信小游戏为什么火了?
    黄金代理这么多,怎么选?
    网络编程01
    Worthington公司精氨酸酶:L-鸟氨酸的制备应用
    轨道姿态常用编程缩写
    Sentinel: 分布式系统的流量防卫兵
  • 原文地址:https://blog.csdn.net/qq_63819197/article/details/133872554