leetcode-37：解数独

• 题目
• 解题
• • 方法一：回溯

题目链接
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：

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：

由于每个数独的长、宽、值都可以用for循环可以遍历的，但是需要填的空格数却不知道，这时候就需要用到回溯，需要填的空格就是递归的次数。 (否则就要用使用 空格个 for循环，显然是不可能的。)

参考链接

class Solution {
public:
    bool backtracing(vector>& board){
        for(int i=0;i>& board){
        for(int i=0;i<9;i++){   // 判断列里是否重复
            if(board[i][col]==val) return false;
        }
        for(int j=0;j<9;j++){   // 判断行里是否重复
            if(board[row][j]==val) return false;
        }
        int startRow=(row/3)*3;
        int startCol=(col/3)*3;
        for(int i=startRow;i>& board){
        backtracing(board);
    }
};