From 1aab307ed7a0e9c539bc450fdf7fbdc3082a1b67 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Sat, 14 Aug 2021 23:18:15 +0800 Subject: [PATCH 1/5] =?UTF-8?q?=E6=9B=B4=E6=96=B0=20=E9=9D=A2=E8=AF=95?= =?UTF-8?q?=E9=A2=9802.07=20=E9=93=BE=E8=A1=A8=E7=9B=B8=E4=BA=A4python?= =?UTF-8?q?=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 用更简单的逻辑完成这道题。 --- problems/面试题02.07.链表相交.md | 37 ++++++++----------------- 1 file changed, 12 insertions(+), 25 deletions(-) diff --git a/problems/面试题02.07.链表相交.md b/problems/面试题02.07.链表相交.md index 9acda71c..8c3a5831 100644 --- a/problems/面试题02.07.链表相交.md +++ b/problems/面试题02.07.链表相交.md @@ -160,34 +160,21 @@ Python： class Solution: def getIntersectionNode(self, headA: ListNode, headB: ListNode) -> ListNode: - lengthA,lengthB = 0,0 - curA,curB = headA,headB - while(curA!=None): #求链表A的长度 - curA = curA.next - lengthA +=1 - - while(curB!=None): #求链表B的长度 - curB = curB.next - lengthB +=1 - - curA, curB = headA, headB + """ + 根据快慢法则，走的快的一定会追上走得慢的。 + 在这道题里，有的链表短，他走完了就去走另一条链表，我们可以理解为走的快的指针。 - if lengthB>lengthA: #让curA为最长链表的头，lenA为其长度 - lengthA, lengthB = lengthB, lengthA - curA, curB = curB, curA + 那么，只要其中一个链表走完了，就去走另一条链表的路。如果有交点，他们最终一定会在同一个 + 位置相遇 + """ + cur_a, cur_b = headA, headB # 用两个指针代替a和b - gap = lengthA - lengthB #求长度差 - while(gap!=0): - curA = curA.next #让curA和curB在同一起点上 - gap -= 1 - while(curA!=None): - if curA == curB: - return curA - else: - curA = curA.next - curB = curB.next - return None + while cur_a != cur_b: + cur_a = cur_a.next if cur_a else headB # 如果a走完了，那么就切换到b走 + cur_b = cur_b.next if cur_b else headA # 同理，b走完了就切换到a + + return cur_a ``` Go： From d350a4dd8206f8cae4773d96e0a985e954276fe6 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Mon, 16 Aug 2021 15:42:28 +0800 Subject: [PATCH 2/5] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200349.=E4=B8=A4?= =?UTF-8?q?=E4=B8=AA=E6=95=B0=E7=BB=84=E7=9A=84=E4=BA=A4=E9=9B=86.md=20pyt?= =?UTF-8?q?hon=E4=BB=A3=E7=A0=81=E7=AE=80=E5=8C=96?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 一行代码解决 --- problems/0349.两个数组的交集.md | 8 +------- 1 file changed, 1 insertion(+), 7 deletions(-) diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 29c1c144..2dd52a85 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -121,13 +121,7 @@ Python： ```python class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: - result_set = set() - - set1 = set(nums1) - for num in nums2: - if num in set1: - result_set.add(num) # set1里出现的nums2元素 存放到结果 - return list(result_set) + return list(set(nums1) & set(nums2)) # 两个数组线变成集合，求交集后还原为数组 ``` From 46c1084f69621055c9fc862cd61042194a6db4c0 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Mon, 16 Aug 2021 15:43:09 +0800 Subject: [PATCH 3/5] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200349.=E4=B8=A4?= =?UTF-8?q?=E4=B8=AA=E6=95=B0=E7=BB=84=E7=9A=84=E4=BA=A4=E9=9B=86.md=20?= =?UTF-8?q?=E9=94=99=E5=88=AB=E5=AD=97=E4=BF=AE=E6=AD=A3?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0349.两个数组的交集.md | 2 +- 1 file changed, 1 insertion(+), 1 deletion(-) diff --git a/problems/0349.两个数组的交集.md b/problems/0349.两个数组的交集.md index 2dd52a85..7489352d 100644 --- a/problems/0349.两个数组的交集.md +++ b/problems/0349.两个数组的交集.md @@ -121,7 +121,7 @@ Python： ```python class Solution: def intersection(self, nums1: List[int], nums2: List[int]) -> List[int]: - return list(set(nums1) & set(nums2)) # 两个数组线变成集合，求交集后还原为数组 + return list(set(nums1) & set(nums2)) # 两个数组先变成集合，求交集后还原为数组 ``` From fbeb80f14ad127e1d47c549c1c2063715896a6bf Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Mon, 16 Aug 2021 18:24:43 +0800 Subject: [PATCH 4/5] =?UTF-8?q?=E6=9B=B4=E6=96=B0=200202.=E5=BF=AB?= =?UTF-8?q?=E4=B9=90=E6=95=B0.md=20python=E4=BB=A3=E7=A0=81=E7=AE=80?= =?UTF-8?q?=E5=8C=96?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 利用闭包的特性，修改了原本python代码中不符合PEP3的写法，以及赘余的代码。思路保持一致 --- problems/0202.快乐数.md | 36 ++++++++++++++++++++---------------- 1 file changed, 20 insertions(+), 16 deletions(-) diff --git a/problems/0202.快乐数.md b/problems/0202.快乐数.md index 1c630b6a..2e784e6a 100644 --- a/problems/0202.快乐数.md +++ b/problems/0202.快乐数.md @@ -111,25 +111,29 @@ Python： ```python class Solution: def isHappy(self, n: int) -> bool: - set_ = set() - while 1: - sum_ = self.getSum(n) - if sum_ == 1: + def calculate_happy(num): + sum_ = 0 + + # 从个位开始依次取，平方求和 + while num: + sum_ += (num % 10) ** 2 + num = num // 10 + return sum_ + + # 记录中间结果 + record = set() + + while True: + n = calculate_happy(n) + if n == 1: return True - #如果这个sum曾经出现过，说明已经陷入了无限循环了，立刻return false - if sum_ in set_: + + # 如果中间结果重复出现，说明陷入死循环了，该数不是快乐数 + if n in record: return False else: - set_.add(sum_) - n = sum_ - - #取数值各个位上的单数之和 - def getSum(self, n): - sum_ = 0 - while n > 0: - sum_ += (n%10) * (n%10) - n //= 10 - return sum_ + record.add(n) + ``` Go： From 175f1c5e1a459e72511ae56ce33828a8fc367690 Mon Sep 17 00:00:00 2001 From: Eyjan_Huang <81480748+Eyjan-Huang@users.noreply.github.com> Date: Mon, 16 Aug 2021 19:11:16 +0800 Subject: [PATCH 5/5] =?UTF-8?q?=E7=AE=80=E5=8C=96=200001.=E4=B8=A4?= =?UTF-8?q?=E6=95=B0=E4=B9=8B=E5=92=8C.md=20python=E4=BB=A3=E7=A0=81?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit 原代码过于赘余，需要遍历两次且可读性较差。更新后的代码在维持题主题意的基础上的一次优化 --- problems/0001.两数之和.md | 15 ++++++++------- 1 file changed, 8 insertions(+), 7 deletions(-) diff --git a/problems/0001.两数之和.md b/problems/0001.两数之和.md index f8c9da5f..9b961d0b 100644 --- a/problems/0001.两数之和.md +++ b/problems/0001.两数之和.md @@ -110,13 +110,14 @@ Python： ```python class Solution: def twoSum(self, nums: List[int], target: int) -> List[int]: - hashmap={} - for ind,num in enumerate(nums): - hashmap[num] = ind - for i,num in enumerate(nums): - j = hashmap.get(target - num) - if j is not None and i!=j: - return [i,j] + records = dict() + + # 用枚举更方便，就不需要通过索引再去取当前位置的值 + for idx, val in enumerate(nums): + if target - val not in records: + records[val] = idx + else: + return [records[target - val], idx] # 如果存在就返回字典记录索引和当前索引 ```