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：
- 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
- 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：
- 给定的数独序列只包含数字 1-9 和字符 '.'
- 你可以假设给定的数独只有唯一解
- 给定数独永远是 9x9 形式的
思路
棋盘搜索问题可以使用回溯法暴力搜索，只不过这次我们要做的是「二维递归」。
回溯三部曲
1.递归函数以及参数
因为解数独找到一个符合的条件（就在树的叶子节点上）立刻就返回，相当于找从根节点到叶子节点一条唯一路径，所以需要使用bool返回值。
2.递归终止条件
本题递归不用终止条件，解数独是要遍历整个树形结构寻找可能的叶子节点就立刻返。
递归的下一层的棋盘一定比上一层的棋盘多一个数，等数填满了棋盘自然就终止。
3.递归单层搜索逻辑
一个for循环遍历棋盘的行，一个for循环遍历棋盘的列，一行一列确定下来之后，递归遍历这个位置放9个数字的可能性。
判断棋盘是否合法
判断棋盘是否合法有如下三个维度：
- 同行是否重复
- 同列是否重复
- 9宫格里是否重复
代码
class Solution {
public void solveSudoku(char[][] board) {
backtracking(board);
}
boolean backtracking(char[][] board) {
for (int i = 0; i < board.length; i++) { // 遍历行
for (int j = 0; j < board[0].length; j++) { // 遍历列
if (board[i][j] != '.') continue;
for (char k = '1'; k <= '9'; k++) { // (i, j) 这个位置放k是否合适
if (isValid(i, j, k, board)) {
board[i][j] = k; // 放置k
if (backtracking(board)) return true; // 如果找到合适一组立刻返回
board[i][j] = '.'; // 回溯，撤销k
}
}
return false; // 9个数都试完了，都不行，那么就返回false
}
}
return true; // 遍历完没有返回false，说明找到了合适棋盘位置了
}
boolean isValid(int row, int col, char val, char[][] board) {
for (int i = 0; i < 9; i++) { // 判断行里是否重复
if (board[row][i] == val) {
return false;
}
}
for (int j = 0; j < 9; j++) { // 判断列里是否重复
if (board[j][col] == val) {
return false;
}
}
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;
}
}
标签：解数,递归,算法,遍历,board,回溯,棋盘,数独 来源： https://www.cnblogs.com/luedong/p/14808298.html
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