链表专题训练

1、合并两个有序链表

  递归：判断两个节点值大小并递归下一次，递归出口为当节点为空时

class Solution:
    def mergeTwoLists(self, list1: Optional[ListNode], list2: Optional[ListNode]) -> Optional[ListNode]:
        if not list1:
            return list2
        if not list2:
            return list1
        if list1.val<list2.val:
            list1.next=self.mergeTwoLists(list1.next,list2)
            return list1
        if list1.val>list2.val:
            list2.next=self.mergeTwoLists(list2.next,list1)
            return list2
    

2、两数相加

class Solution:
    def addTwoNumbers(self, l1: Optional[ListNode], l2: Optional[ListNode]) -> Optional[ListNode]:
        pre = ListNode(-1)
        curr = pre
        carry = 0
        while(l1 != None or l2 != None):
            x = 0 if l1 == None else l1.val
            y = 0 if l2 == None else l2.val
            _sum = int(x + y + carry)
            carry = 1 if _sum > 9 else 0
            _sum = _sum % 10
            curr.next = ListNode(_sum)
            curr = curr.next
            if l1 != None:
                l1 = l1.next
            if l2 != None:
                l2 = l2.next
        if carry == 1:
            curr.next = ListNode(1)
        return pre.next

标签：None,专题,ListNode,训练,next,链表,l2,l1,Optional
来源： https://www.cnblogs.com/feiyuyu/p/16512855.html