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  • Python两数相加

    两数相加

    给出两个 非空 的链表用来表示两个非负的整数。其中,它们各自的位数是按照 逆序 的方式存储的,并且它们的每个节点只能存储 一位 数字。

    如果,我们将这两个数相加起来,则会返回一个新的链表来表示它们的和。

    您可以假设除了数字 0 之外,这两个数都不会以 0 开头。

    示例:

    输入:(2 -> 4 -> 3) + (5 -> 6 -> 4)
    输出:7 -> 0 -> 8
    原因:342 + 465 = 807
    

    思路一:

    # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None
    
    class Solution:
        def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
            lst1 = []
            while l1:
                lst1.append(l1.val)
                l1 = l1.next
            lst2 = []
            while l2:
                lst2.append(l2.val)
                l2 = l2.next
            m, n = len(lst1), len(lst2)
            if m < n:
                m, n, lst1, lst2 = n, m, lst2, lst1
            lst3 = []
            for i in range(0, m):
                if i < n:
                    num = lst1[i] + lst2[i]
                else:
                    num = lst1[i]
                if num > 9:
                    num = num - 10
                    lst3.append(num)
                    if i == m - 1:
                        lst3.append(1)
                    else:
                        lst1[i + 1] += 1
                else:
                    lst3.append(num)
            node = ListNode(lst3[0])
            cur = node
            for i in lst3[1:]:
                node_ = ListNode(i)
                cur.next = node_
                cur = node_
            return node
    

    思路二:

    # Definition for singly-linked list.
    # class ListNode:
    #     def __init__(self, x):
    #         self.val = x
    #         self.next = None
    
    class Solution:
        def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
            jin_wei = 0
            head = cur = None
            while l1 or l2:
                l1_val = l1.val if l1 else 0
                l2_val = l2.val if l2 else 0
                l1 = l1.next if l1 else None
                l2 = l2.next if l2 else  None
                num = l1_val + l2_val
                if jin_wei == 1:
                    num += 1
                    jin_wei = 0
                if num > 9:
                    num = num - 10
                    jin_wei = 1
                if not head:
                    head = ListNode(num)
                    cur = head
                else:
                    node = ListNode(num)
                    cur.next = node
                    cur = node
            if jin_wei == 1:
                cur.next = ListNode(1)
            return head
    
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  • 原文地址:https://www.cnblogs.com/zyyhxbs/p/12771778.html
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