LeetCode---【和的操作】

两数之和

我的答案

class Solution(object):def twoSum(self, nums, target):leng=len(nums)for i in range(leng):for j in range(leng):if nums[i]+nums[j]==target and i!=j:return [i,j]

在b站up那里学到的【然后自己复写】

• 忘记怎么取list的第二个值了。【利用range的特性】

class Solution(object):def twoSum(self, nums, target):leng=len(nums)for i in range(0,leng):for j in range(i+1,leng):if nums[i]+nums[j]==target and i!=j:return [i,j]

• 哈希表【空间换时间】

class Solution(object):def twoSum(self, nums, target):mapping = {}for i in range(0, len(nums)):mapping[nums[i]] = ifor j in range(0, len(nums)):diff = target - nums[j]if (diff in mapping and mapping[diff] != j):return [j,mapping[diff]];

和为 K 的子数组

在b站up那里学到的【然后自己复写】

class Solution(object):def subarraySum(self, nums, k):count = 0dic={}dic[0] = 1cur_sum = 0for num in nums:cur_sum = cur_sum + numif cur_sum-k in dic:# 很巧的思路count=count+dic[cur_sum-k]if cur_sum in dic:dic[cur_sum] +=1else:dic[cur_sum] = 1return count        

三数之和

在b站up那里学到的【然后自己复写】

class Solution(object):def threeSum(self, nums):resol=[]nums.sort()for i in range(len(nums)):cur = iif nums[cur]>0:return resolif cur>0 and nums[cur-1] == nums[cur]:continue # 去重left = cur+1right = len(nums)-1while left<right:total = nums[cur]+nums[left]+nums[right]if total >0:right-=1elif total<0:left+=1else:resol.append([nums[cur],nums[left],nums[right]])while left<right and nums[right-1]==nums[right]:# 去重right-=1while left<right and nums[left+1]==nums[left]:# 去重left+=1right-=1left+=1               return resol

两数相加【链表】

我的半路答案：没有看到是链表

• 我还没有写完，只是想半路测一下是否正确。
• 关于链表的话，我相对还是弱一点，下次巩固。

class Solution(object):def addTwoNumbers(self, l1, l2):resl=[]cur=0len1=len(l1)len2=len(l2)for i in range(min(len1,len2)):total=l1[i]+l2[i]+curif total>10:cur=total/10resl.append(total%10)else:resl.append(total)return resl

在b站up那里学到的【复写失败后整理】

class Solution(object):def addTwoNumbers(self, l1, l2):total = 0next1 = 0dummy = ListNode()cur = dummy#############必须得要一个临时变量，因为后面输出是从头开始输出while (l1 != None and l2 != None):total = l1.val + l2.val + next1cur.next = ListNode(total % 10)next1 = total // 10cur = cur.next##################不要忘记l1 = l1.nextl2 = l2.nextwhile l1 != None:total = l1.val + next1cur.next = ListNode(total % 10)next1 = total // 10cur = cur.next##################不要忘记l1 = l1.nextwhile l2 != None:total = l2.val + next1cur.next = ListNode(total % 10)next1 = total // 10cur = cur.next##################不要忘记l2 = l2.nextif next1 != 0:cur.next = ListNode(next1)return dummy.next