37. Sudoku Solver

题目

Write a program to solve a Sudoku puzzle by filling the empty cells.

A sudoku solution must satisfy all of the following rules:

Each of the digits 1-9 must occur exactly once in each row.
Each of the digits 1-9 must occur exactly once in each column.
Each of the the digits 1-9 must occur exactly once in each of the 9 3x3 sub-boxes of the grid.

Empty cells are indicated by the character '.'.

A sudoku puzzle...

...and its solution numbers marked in red.

Note:

The given board contain only digits 1-9 and the character '.'.
You may assume that the given Sudoku puzzle will have a single unique solution.
The given board size is always 9x9.

题目大意

编写一个程序，通过已填充的空格来解决数独问题。一个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。

空白格用 '.' 表示。

解题思路

给出一个数独谜题，要求解出这个数独
解题思路 DFS 暴力回溯枚举。数独要求每横行，每竖行，每九宫格内，1-9 的数字不能重复，每次放下一个数字的时候，在这 3 个地方都需要判断一次。
另外找到一组解以后就不需要再继续回溯了，直接返回即可。

代码


package leetcode

type position struct {
	x int
	y int
}

func solveSudoku(board [][]byte) {
	pos, find := []position{}, false
	for i := 0; i < len(board); i++ {
		for j := 0; j < len(board[0]); j++ {
			if board[i][j] == '.' {
				pos = append(pos, position{x: i, y: j})
			}
		}
	}
	putSudoku(&board, pos, 0, &find)
}

func putSudoku(board *[][]byte, pos []position, index int, succ *bool) {
	if *succ == true {
		return
	}
	if index == len(pos) {
		*succ = true
		return
	}
	for i := 1; i < 10; i++ {
		if checkSudoku(board, pos[index], i) && !*succ {
			(*board)[pos[index].x][pos[index].y] = byte(i) + '0'
			putSudoku(board, pos, index+1, succ)
			if *succ == true {
				return
			}
			(*board)[pos[index].x][pos[index].y] = '.'
		}
	}
}

func checkSudoku(board *[][]byte, pos position, val int) bool {
	// 判断横行是否有重复数字
	for i := 0; i < len((*board)[0]); i++ {
		if (*board)[pos.x][i] != '.' && int((*board)[pos.x][i]-'0') == val {
			return false
		}
	}
	// 判断竖行是否有重复数字
	for i := 0; i < len((*board)); i++ {
		if (*board)[i][pos.y] != '.' && int((*board)[i][pos.y]-'0') == val {
			return false
		}
	}
	// 判断九宫格是否有重复数字
	posx, posy := pos.x-pos.x%3, pos.y-pos.y%3
	for i := posx; i < posx+3; i++ {
		for j := posy; j < posy+3; j++ {
			if (*board)[i][j] != '.' && int((*board)[i][j]-'0') == val {
				return false
			}
		}
	}
	return true
}