【力扣】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 输入数独仅有一个解

题解：

回溯法：

func solveSudoku(board [][]byte) {
	var isValid func(row, col int, k byte, board [][]byte) bool
	isValid = func(row, col int, k byte, board [][]byte) bool {
		// 检查行里面是否有重复的数字
		for i := 0; i < 9; i++ {
			if board[row][i] == k {
				return false
			}
		}
		// 检查列里面是否有重复的数字
		for i := 0; i < 9; i++ {
			if board[i][col] == k {
				return false
			}
		}
		// 检查九宫格内是否有重复的数字
		startRow := row / 3 * 3
		startCol := col / 3 * 3
		for i := startRow; i < startRow+3; i++ {
			for j := startCol; j < startCol+3; j++ {
				if board[i][j] == k {
					return false
				}
			}
		}
		return true
	}
	var backtracking func() bool
	backtracking = func() bool {
		for i := 0; i < len(board); i++ {
			for j := 0; j < len(board[0]); j++ {
				// 判断该位置是否适合填数字
				if board[i][j] != '.' {
					continue
				}
				// 数字 1-9 逐个尝试
				for k := '1'; k <= '9'; k++ {
					if isValid(i, j, byte(k), board) {
						board[i][j] = byte(k)
						if backtracking() {
							return true
						}
						// 回溯
						board[i][j] = '.'
					}
				}
				return false
			}
		}
		return true
	}
	backtracking()
}