数据结构与算法--回溯法

回溯法

1 括号生成
- 分析：
2 解数独
- 分析
- 代码

回溯法本质是的暴力枚举/遍历法，一般用递归实现。
当我们可以把问题分解为若干个步骤，每个步骤都有若干个选择的时候，若需要列出所有解答形式，则采用枚举法。

1 括号生成

数字 n 代表生成括号的对数，请你设计一个函数，用于能够生成所有可能的并且有效的括号组合。

分析：

有效的括号一定是现有左括号，再有有括号，无论在哪个位置，左括号的数量要大于等于右括号。

n=1时：

1个（，一个），第一个位置永远只能是（，第二个位置就只剩下）了,所以只有["()"]

n=2时：

第一个只能是(,第二个可以是(也可以是)，
第二个是(时，两个(都用完了，第三个和第四个都只能是)了
第二个是）时，目前‘（’个数等于‘)’个数，第三个只能是‘（’，第三个和第四个都只能是)`了

pos	0	1	2	3
	(	（	）	）
	(	）	（	）

综上，我们需要num_left,num_right计算目前已经有的左右括号的个数，只有右括号个数小于左括号个数的时候，下一个才能是右括号，否则（右括号个数等于左括号个数）下一个只能左括号，可以用递归的形式写。

#python
class Solution:def generateParenthesis(self, n: int) -> list[str]:def helper(index,n,right_num,left_num):if len(subs)==2*n:ans.append(''.join(subs))elif index<2*n:if index==0:subs.append('(')helper(index+1,n,right_num,left_num+1)else:if left_num<n:subs.append('(')helper(index+1,n,right_num,left_num+1)subs.pop()if right_num<left_num:subs.append(')')helper(index+1,n,right_num+1,left_num)subs.pop()subs = []ans = []helper(0,n,0,0)return ans

2 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。（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。
在这里插入图片描述

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：

在这里插入图片描述

分析

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。

可以遍历每一个数字，找不到的时候回溯,分析：
使用（i,j）表示当前遍历的位置，使用row[i]表示第i行已经存在的数字，col[j]表示第j列已经存在的数字，box[(i//3)*3+j//3]是当前位置所在格子已经存在的数，row[i]，col[j]，box[(i//3)*3+j//3]是哈希表方便查询。
边界条件：

当i==9的时候，代表所有格子都遍历完了
当j==9的时候，代表要换行了
当board[i][j]为数字时，不用填数字，直接下一个

非边界条件：
这个时候要从1-9里面选还没有在row[i]，col[j]，box[(i//3)*3+j//3]中出现的数字填进格子，如果helper(i,j+1) return true,说明可以填；否则，回溯，删掉数字继续遍历，如果所有数字都遍历完也没找到，return False

代码

class Solution:def solveSudoku(self, board: list[list[str]]) -> None:"""Do not return anything, modify board in-place instead."""def helper(board,row,col,box,i,j):if i==9:return Trueif j==9:return helper(board,row,col,box,i+1,0)if board[i][j]!='.':return helper(board,row,col,box,i,j+1)box_index = (i//3)*3+(j//3)for num in range(1,10):if num not in row[i] and num not in col[j] and num not in box[box_index]:row[i][num]=1col[j][num]=1box[box_index][num]=1board[i][j]=str(num)if helper(board,row,col,box,i,j+1):return Truedel row[i][num]del col[j][num]del box[box_index][num]board[i][j]='.'return Falsem = len(board)n = len(board[0])row = [{} for _ in range(m)]col = [{} for _ in range(n)]box = [{} for _ in range(m // 3 * n // 3)]for i in range(m):for j in range(n):num = board[i][j]if num!='.':num = int(num)row[i][num]=1col[j][num]=1box[(i//3)*3+(j//3)][num]=1helper(board,row,col,box,0,0)return board