Lee Code 学习 python

最新推荐文章于 2025-07-10 15:58:08 发布

转载最新推荐文章于 2025-07-10 15:58:08 发布 · 357 阅读

python 专栏收录该内容

1 篇文章

订阅专栏

本文深入探讨了两个核心算法：两数之和与链表相加。首先介绍了如何寻找数组中两个数使它们的和等于特定目标值的方法，随后详细解析了链表相加算法，展示了如何处理链表节点的相加及进位问题。文章通过具体实例和代码实现，为读者提供了清晰的算法理解和实践指导。

TWO NUM

class Solution:
    # @return a tuple, (index1, index2)
    def twoSum(self, num, target):
        dict = {}
        for i in range(len(num)):
            x = num[i]
            if target-x in dict:
                return (dict[target-x], i)
            dict[x] = i


print(Solution().twoSum(num = [0,2,7,11,15],target=9))

ADD TWO NUM

import math
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None

class Solution:
    def addTwoNumbers(self, l1, l2):
        """
        :type l1: ListNode
        :type l2: ListNode
        :rtype: ListNode
        """
        resultNode = None
        add= 0
        while True:
            if l1.val == -1:
                l1.val =0
            if l2.val ==-1:
                l2.val = 0

            tSum = (l1.val+l2.val+add)%10
            add = math.floor((l1.val+l2.val+add)/10)
            listn = ListNode(tSum)

            if resultNode ==None:
                resultNode =listn
                flagNode = resultNode
            else:
                flagNode.next =listn
                flagNode = flagNode.next

            if l1.next != None:
                l1 = l1.next
            else:
                l1.val = -1

            if l2.next != None:
                l2 = l2.next
            else:
                l2.val = -1

            if l1.val ==-1 and l2.val ==-1:
                break

        if add != 0:
                listn = ListNode(add)
                flagNode.next =listn

        return resultNode