1. two sum

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class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
      # xa + xb = target,  xb = target - xa,  use dictionary {} to store key-value pair
      # 用dictionary来存key-value对，如果key= target-nums[i] 在字典中则返回 [dict[key], i]
      # 否则，把当前的值 nums[i] 做key 存入字典 dict[nums[i]] = i 用做以后的比较
        dict = {}
        for i, nums[i] in enumerate(nums):
            key = target - nums[i]
            if key in dict:
                return [dict[key], i]
            else:
                dict[nums[i]] = i

每日一记：

创始人刷题的经验:刷题三千多，很感动人。我翻了一下知乎上关于刷题班的评价，众说纷纭。但是，对于刷题这件事，即使是senior级别面试，也是绕不过去的。那么我们就重新开始刷题和训练system design吧，不要让自己闲下来。闲下来也就是刷微信，刷剧，对于我们的成长并无帮助。

21. Merge Two Sorted Lists

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# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    # Time complexity : O(n+m) ; Space complexity : O(1)
     # maintain an unchanging reference to node ahead of the return node.
    def mergeTwoLists(self, l1: ListNode, l2: ListNode) -> ListNode:
        prev_head = ListNode(0)
        prev = prev_head
        
        while l1 != None and l2 != None:
            if l1.val <= l2.val:
                prev.next = l1
                l1 = l1.next
            else:
                prev.next = l2
                l2 = l2.next
            prev = prev.next
        
        if l1 == None :
            prev.next = l2
        if l2 == None :
            prev.next = l1
        
        return prev_head.next