力扣61~65题
题61（中等）：
分析：
python代码：
# Definition for singly-linked list.
# class ListNode:
# def __init__(self, val=0, next=None):
# self.val = val
# self.next = next
class Solution:def rotateRight(self, head: Optional[ListNode], k: int) -> Optional[ListNode]:#利用空间复杂度解决，将链表装进列表node_list=[]p=headwhile p!=None:node_list.append(p)p=p.nextnode_len=len(node_list)if node_len==0:return headk_new=k%node_lenif k_new==0:return headnode_list[-1].next=node_list[0]head=node_list[-k_new]node_list[-(k_new+1)].next=Nonereturn head题62（中等）：
分析：
python代码：
class Solution:def uniquePaths(self, m: int, n: int) -> int:#思路就是动态数组嘛auto_list=[[1 for j in range(n)] for i in range(m)]for i in range(1,m):for j in range(1,n):a1,a2=0,0if i-1>=0:a1=auto_list[i-1][j]if j-1>=0:a2=auto_list[i][j-1]auto_list[i][j]=a1+a2return auto_list[-1][-1]题63（中等）：
分析
会这个差不多可以制作迷宫小游戏了
python代码：
class Solution:def uniquePathsWithObstacles(self, obstacleGrid: List[List[int]]) -> int:#动态数组，和上一题差不多#构造动态数组，obstacleGrid中有障碍为设置为0，没有设置为1auto_list=[[0 for i in range(len(obstacleGrid[0]))] for j in range(len(obstacleGrid))]for i in range(len(obstacleGrid)):for j in range(len(obstacleGrid[0])):if obstacleGrid[i][j]==0:auto_list[i][j]=1else:auto_list[i][j]=0#设置动态数组的值for i in range(len(obstacleGrid)):for j in range(len(obstacleGrid[0])):#第一个位置必须是0if i==0 and j==0:continueif auto_list[i][j]==0:continuea1,a2=0,0if i-1>=0:a1=auto_list[i-1][j]if j-1>=0:a2=auto_list[i][j-1]auto_list[i][j]=a1+a2return auto_list[-1][-1]题64（中等）：
分析：
python代码：
class Solution:def minPathSum(self, grid: List[List[int]]) -> int:#动态规划，本质上和前面的题目是一样的auto_list=grid.copy()for i in range(len(auto_list)):for j in range(len(auto_list[0])):if i==0 and j==0:#第一个就跳过就好了continueif i-1<0:auto_list[i][j]+=auto_list[i][j-1]continueif j-1<0:auto_list[i][j]+=auto_list[i-1][j]continueauto_list[i][j]+=min(auto_list[i-1][j],auto_list[i][j-1])return auto_list[-1][-1]题65（困难）：
分析：
不得不说，这个有效数字判定规则很诡异，不过难度真不配困难
python代码：
class Solution:def isNumber(self, s: str) -> bool:s_stack=[]dot_flag=0sign_flag=0e_flag=0for i in s:#如果为数字if i in ('0','1','2','3','4','5','6','7','8','9'):if dot_flag==1:dot_flag=0if e_flag==1:e_flag=0if sign_flag==1:sign_flag=0s_stack.append(i)#如果为.elif i=='.':if s_stack==[] or s_stack==['+'] or s_stack==['-']:dot_flag=1if '.' in s_stack:return Falseif ('e' in s_stack) or ('E' in s_stack):return Falses_stack.append('.')#如果为+-elif i=='+' or i=='-':#如果不是空或者前面不是e就0if s_stack==[]:sign_flag=1s_stack.append(i)elif s_stack[-1]=='e' or s_stack[-1]=='E':s_stack.append(i)else:return False#如果为e或者Eelif i=='e' or i=='E':print(s_stack)e_flag=1if 'e' in s_stack or 'E' in s_stack:return Falseif s_stack==[] or s_stack==['+'] or s_stack==['-']:return Falseif s_stack==['.'] or s_stack==['+','.'] or s_stack==['-','.']:return Falses_stack.append(i)else:return Falseif dot_flag==1:return Falseif e_flag==1:return Falseif sign_flag==1:return Falsereturn True