leetcode 链表求和 两种方法 迭代和递归

面试题 02.05. 链表求和

https://leetcode-cn.com/problems/sum-lists-lcci/

给定两个用链表表示的整数，每个节点包含一个数位。这些数位是反向存放的，也就是个位排在链表首部。编写函数对这两个整数求和，并用链表形式返回结果。

示例：

输入：(7 -> 1 -> 6) + (5 -> 9 -> 2)，即617 + 295
输出：2 -> 1 -> 9，即912

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def addTwoNumbers(self, l1, l2):

        head = ListNode(0)
        curr = head
        temp = 0

        while l1 and l2: # l1 and l2 are not None
            numsum = l1.val + l2.val + temp
            temp = numsum // 10
            curr.next = ListNode(numsum % 10)
            curr = curr.next
            l1, l2 = l1.next, l2.next


        # l1 or l2 is None
        rest = l1 if l1 else l2
        
        while rest and temp:
            numsum = temp + rest.val
            temp = numsum // 10
            curr.next = ListNode(numsum % 10)
            curr = curr.next
            rest = rest.next


        if not temp and not rest: 
            # temp and rest are both None
            pass    
        elif not temp: 
            # rest still exist
            curr.next = rest 
        elif not rest: 
            # temp is not 0
            curr.next = ListNode(temp)
        return head.next

结果还凑活吧

这道题主要是分三步:

第一步:两个链表都不为空的时候, 那就是三者直接相加, 之和不会超过19, 所以取余%10 为当前位的数值, 整除10 为进位

第二步:当两个链表至少有一个为空时, 先假设有一个链表存在, 将其与进位进行继续之前的操作, 只是现在是两个相加

第三步:判断第二步终止的情况, 处理结尾

直接返回链表的头即可

 

第二种方法 递归

class Solution:
    def addTwoNumbers(self, l1: ListNode, l2: ListNode) -> ListNode:
        return self.addTwoNumbers2(l1, l2, 0)

    def addTwoNumbers2(self, l1, l2, z):
        if l1 is None and l2 is None and z == 0:
            return None

        if l1 is None:
            l1 = ListNode(0)

        if l2 is None:
            l2 = ListNode(0)

        value = l1.val + l2.val + z
        result = ListNode(value % 10)
        result.next = self.addTwoNumbers2(l1.next, l2.next, value//10)
        return result

效率也还可以: