classSolution(object):deftwoSum(self, numbers, target):""":type numbers: List[int] :type target: int :rtype: List[int]"""ifnotnumbers:return[]foriinxrange(len(numbers)):ifi > 0andnumbers[i-1] ==numbers[i]:continueind1=i ind2= len(numbers) - i -1whileind1 <=ind2: sumv= ...
# we cut of the linked list at middle slow.next = None left = self.sortList(head) right = self.sortList(head2) return self.merge(left, right) 主要是在merge的时候。要推断第一个结点 AC代码 class Solution: def merge(self, head1, head2): if head1 == None: return head2 if head2 ...
classSolution:deftwoSum(self,nums,target):""":type nums: List[int]:type target: int:rtype: List[int]"""#用len()方法取得nums列表的长度n=len(nums)#x取值从0一直到n(不包括n)forxinrange(n):#y取值从x+1一直到n(不包括n)#用x+1是减少不必要的循环,y的取值肯定是比x大foryinrange(x+1...
from typing import List class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: for i in range(len(nums)): for j in range(len(nums)): if target == nums[i] + nums[j] and i != j: return [i, j] else: continue return None print(Solution().twoSum([...
class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: #定义...
避免生成重复的答案(跳过 'num[i] == num[i - 1]' ),也可以利用Python的特性,将答案转换为 set 再转换为list来消除重复的答案 注意索引的范围 我初次编写的时候采用比较容易想到的解法方案,可以打败50.3%的其他代码,如下: classSolution:deffourSum(self,nums,target):""" ...
对应的 Java 仓库的地址,传送门:https://github.com/liweiwei1419/LeetCode-Solution-Java 说明:现在刷题,尤其是写题解,绝大多数问题都会写两个语言的代码,Java 是我的母语,Python 是我的新欢。 发布在 LeetCode 中文版上的题解配图使用的 PPT,传送门:https://github.com/liweiwei1419/LeetCode-Solution-PPT...
classSolution: defsetZeroes(self,matrix:List[List[int]])->None: m,n=len(matrix),len(matrix[0]) row,col=[False]*m, [False]*n foriinrange(m): forjinrange(n): ifmatrix[i][j]==0: row[i]=col[j]=True foriinrange(m): ...
ifisinstance(l1,list): l1 =ListNode(l1)l2 =ListNode(l2) 我们来看看 vscode 调试打印的效果: if__name__ =="__main__":test= Solution()print(test.addTwoNumbers([1,3],[2,1,3])) 调试结果:(和官方定义得那个输出是一样的) f:/Leetcode/2.两数相加.py ListNode {val:3,next: ListNode {...
class Solution(object): def minPathSum(self, grid): """ :type grid: List[List[int]] :rtype: int """ # 解法一, 空间优化 m, n = len(grid), len(grid[0]) for i in range(1, n): # 第一行初始化 grid[0][i] = grid[0][i] + grid[0][i - 1] ...