力扣第37题：解数独
编写一个程序，通过填充空格来解决数独问题。
数独示例图
相关知识点：回溯算法
回溯算法的基本思想是：从一条路往前走，能进则进，不能进则退回来，换一条路再试。
八皇后问题就是回溯算法的典型，第一步按照顺序放一个皇后，然后第二步符合要求放第2个皇后，如果没有位置符合要求，那么就要改变第一个皇后的位置，重新放第2个皇后的位置，直到找到符合条件的位置就可以了。
回溯在迷宫搜索中使用很常见，就是这条路走不通，然后返回前一个路口，继续下一条路。
回溯算法说白了就是穷举法。
不过回溯算法使用剪枝函数，剪去一些不可能到达最终状态（即答案状态）的节点，从而减少状态空间树节点的生成。
回溯法是一个既带有系统性又带有跳跃性的搜索算法。
它在包含问题的所有解的解空间树中，按照深度优先的策略，从根结点出发搜索解空间树。算法搜索至解空间树的任一结点时，总是先判断该结点是否肯定不包含问题的解。如果肯定不包含，则跳过对以该结点为根的子树的系统搜索，逐层向其祖先结点回溯。否则，进入该子树，继续按深度优先的策略进行搜索。回溯法在用来求问题的所有解时，要回溯到根，且根结点的所有子树都已被搜索遍才结束。
而回溯法在用来求问题的任一解时，只要搜索到问题的一个解就可以结束。这种以深度优先的方式系统地搜索问题的解的算法称为回溯法，它适用于解一些组合数较大的问题。
数独的解法也可采用回溯算法。
数独的解法需遵循如下规则：
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。（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：
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
来源：力扣（LeetCode）
参考实现（附C++代码）：
//作者：文方俊
//日期：2021-05-18
//功能：解数独
class Solution {
public:
bool solveSudoku(vector<vector<char>>& board) {
for(int i=0;i<board.size();i++){
for(int j=0;j<board[0].size();j++){
if(board[i][j]!='.'){
continue;
}
for(char ch='1';ch<='9';ch++){
if(isValidBoard(board,i,j,ch)){ //if board is valid
board[i][j]=ch; //try '1'-'9'
if(solveSudoku(board)){
return true;
}
board[i][j]='.';//back to start
}
}
return false;//try again
}
}
return true;
}
bool isValidBoard(vector<vector<char>>&board,int row,int col,char ch){
for(int i=0;i<9;i++){
if(board[row][i]==ch){
return false;
}
}
for(int j=0;j<9;j++){
if(board[j][col]==ch){
return false;
}
}
int rowStart=(row/3)*3;
int colStart=(col/3)*3;
for(int i=rowStart;i<rowStart+3;i++){
for(int j=colStart;j<colStart+3;j++){
if(board[i][j]==ch){
return false;
}
}
}
return true;
}
};
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