R3-链表-链表高精度加法

目录

递归法

迭代 

递归法

 l1.val+l2.val+carry，得到的和，%10为当前位存储的值，除以10为当前的进位值

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode],carry=0) -> Optional[ListNode]:
        #递归边界,如果l1,l2都是空结点
        if l1 is None and l2 is None:
            if carry:
                return ListNode(carry)
            return None
        if l1 is None:
            l1,l2=l2,l1
        s=carry+l1.val+(l2.val if l2 else 0)
        l1.val=s%10
        l1.next=self.addTwoNumbers(l1.next,l2.next if l2 else None,s//10)
        return l1

        

 

迭代 

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode],carry=0) -> Optional[ListNode]:
        #迭代法，感觉有点像模拟这个过程，创建链表，每次保存carry值
        #dummy是哨兵结点
        cur=dummy=ListNode()
        carry=0
        while l1 or l2 or carry:
            s=carry+(l1.val if l1 else 0)+(l2.val if l2 else 0)
            cur.next=ListNode(s%10)
            carry=s//10
            cur=cur.next
            if l1:
                l1=l1.next
            if l2:
                l2=l2.next
        return dummy.next