leetcode100HOT

原创已于 2022-02-24 18:34:06 修改 · 481 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #算法

于 2022-02-22 22:54:27 首次发布

学习专栏收录该内容

1 篇文章

订阅专栏

博主分享了三种不同时间复杂度的解决方案，包括两数之和的迭代和查找优化，以及链表加法问题的实现。讨论了ListNode类在LeetCode中遇到的难点，并提供了逐步解析和自理解析版本。关键词：两数之和、链表操作、时间复杂度优化

本人菜鸟边刷边进步

链接：https://leetcode-cn.com/playground/new/empty/

# first  2204ms
class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for i in range(len(nums)):
            for j in range(i+1,len(nums)):
                if nums[i] + nums[j] == target:
                    return i,j
# second 412ms
class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        for index, num in enumerate(nums):
            if target - num in nums and nums.index(target - num) != index:
                return nums.index(target - num), index
# third  12ms
class Solution(object):
    def twoSum(self, nums, target):
        """
        :type nums: List[int]
        :type target: int
        :rtype: List[int]
        """
        dict_ = {}
        for index, num in enumerate(nums):
            if target - num in dict_:
                return dict_[target - num], index
            dict_[num] = index

20220223

链接 https://leetcode-cn.com/problems/add-two-numbers/
这道题在ListNode()难倒了想了很久没弄懂，看看睡觉前能不能弄懂
下午才弄懂，自己写了个简单版

#         摘抄网友的
# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        head = curr = ListNode()
        carry = val = 0

        while carry or l1 or l2:
            val = carry

            if l1: l1, val = l1.next, l1.val + val
            if l2: l2, val = l2.next, l2.val + val

            carry, val = divmod(val, 10)
            curr.next = curr = ListNode(val)
        
        return head.next

# my 理解 version   64ms
class Solution:  
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        return_node = None
        # 初始化进位的值
        jinwei = 0
        # 将结果保存起来 方便后面做逆序
        result = []
        while l1 or l2 or jinwei:
            # 这里需要将进位的值赋值给val，加上进位的值
            val = jinwei
            if l1:
                # 取出数据，并且l1指向l1.next
                val = l1.val + jinwei
                l1 = l1.next
            if l2:
                # 取出数据，并且l2指向l2.next
                val = val + l2.val
                l2 = l2.next
            # 获得进位
            jinwei, yushu = divmod(val, 10)
            result.append(yushu)
        for ii in reversed(result):
            # 新建ListNode()结点
            new = ListNode()
            new.val = ii
            new.next = return_node
            # 将return_node指向new结点
            return_node = new
        return return_node