每日归档： 2022年1月18日

给定一个正整数 n ，输出外观数列的第 n 项。

「外观数列」是一个整数序列，从数字 1 开始，序列中的每一项都是对前一项的描述。

你可以将其视作是由递归公式定义的数字字符串序列：

• countAndSay(1) = "1"
• countAndSay(n) 是对 countAndSay(n-1) 的描述，然后转换成另一个数字字符串。

前五项如下：

1.     1
2.     11
3.     21
4.     1211
5.     111221
第一项是数字 1 
描述前一项，这个数是 1 即 “ 一 个 1 ”，记作 "11"
描述前一项，这个数是 11 即 “ 二 个 1 ” ，记作 "21"
描述前一项，这个数是 21 即 “ 一 个 2 + 一 个 1 ” ，记作 "1211"
描述前一项，这个数是 1211 即 “ 一 个 1 + 一 个 2 + 二 个 1 ” ，记作 "111221"

要 描述 一个数字字符串，首先要将字符串分割为 最小 数量的组，每个组都由连续的最多 相同字符 组成。然后对于每个组，先描述字符的数量，然后描述字符，形成一个描述组。要将描述转换为数字字符串，先将每组中的字符数量用数字替换，再将所有描述组连接起来。

例如，数字字符串 "3322251" 的描述如下图：

示例 1：

输入：n = 1
输出："1"
解释：这是一个基本样例。

示例 2：

输入：n = 4
输出："1211"
解释：
countAndSay(1) = "1"
countAndSay(2) = 读 "1" = 一 个 1 = "11"
countAndSay(3) = 读 "11" = 二 个 1 = "21"
countAndSay(4) = 读 "21" = 一 个 2 + 一 个 1 = "12" + "11" = "1211"

提示：

• 1 <= n <= 30

代码：

# @lc app=leetcode.cn id=38 lang=python3
#
# [38] 外观数列
#

# @lc code=start
from unittest import result


class Solution:
    def countAndSay(self, n: int) -> str:
      def subAndSay(n):
        if n==1 :
            return "1*"
        string = subAndSay(n-1)
        m=1
        result=""
        for i in range(len(string)-1):
            if string[i]==string[i+1]:
                m=m+1
            else: 
                result+=str(m)+string[i]
                m=1
        result+="*"
        return result
      return subAndSay(n)[:-1]

        

结果：

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：

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

思路：回溯法：通俗理解就是如果board[i][j]=VALUE不满足条件就回退到上一步的选择，重新选择。

回溯法（探索与回溯法）是一种选优搜索法，又称为试探法，按选优条件向前搜索，以达到目标。但当探索到某一步时，发现原先选择并不优或达不到目标，就退回一步重新选择，这种走不通就退回再走的技术为回溯法，而满足回溯条件的某个状态的点称为“回溯点”。

代码实现：（递归）【参考了解题思路】

# [37] 解数独
#
#回溯法
"""
回溯法（探索与回溯法）是一种选优搜索法，又称为试探法，
按选优条件向前搜索，以达到目标。但当探索到某一步时，
发现原先选择并不优或达不到目标，就退回一步重新选择，
这种走不通就退回再走的技术为回溯法，
而满足回溯条件的某个状态的点称为“回溯点”。

"""
# @lc code=start
class Solution:
    def solveSudoku(self, board)->None:
        """
        Do not return anything, modify board in-place instead.
        """
        #判断改行、列、3*3小格子是否满足数独规则：

        def isRowSafe(row,value):
            for i in range(9):
                if board[row][i]==value:
                    return False
            return True
    
        def isColSafe(col,value):
            for i in range(9):
                if board[i][col]==value:
                    return False
            return True
        
        def isSmallboxSafe(row,col,value):
            inirow=row//3*3
            inicol=col//3*3
            for i in range(3):
                for j in range(3):
                    if board[i+inirow][j+inicol]==value:
                        return False  
            return True
        #判断该位置是否可行
        def isSafe(row,col,value):
            return isRowSafe(row,value) and isColSafe(col,value) and isSmallboxSafe(row,col,value)                 
        #解数独，结束条件
        def solve(row,col):
            if row==8 and col ==9:
                return True
            if col ==9:
                col=0
                row+=1
            if board[row][col]!=".":
                return solve(row,col+1)
            for i in range(1,10):
                if isSafe(row,col,str(i)):
                    i=str(i)
                    board[row][col] = i
                    if solve(row, col+1):
                        return board
           #回溯到上一个状态（也就是前一个solve）
            board[row][col]="."
            return False
        solve(0,0)
        print(board)