83. 删除排序链表中的重复元素

为保证权益，题目请参考 83. 删除排序链表中的重复元素(From LeetCode).

解决方案1

CPP

struct ListNode
{
    int val;
    ListNode *next;
    ListNode() : val(0), next(nullptr) {}
    ListNode(int x) : val(x), next(nullptr) {}
    ListNode(int x, ListNode *next) : val(x), next(next) {}
};

class Solution
{
public:
    ListNode *deleteDuplicates(ListNode *head)
    {
        if (head == nullptr)
        {
            return nullptr;
        }

        ListNode *preNode = nullptr;
        ListNode *curNode = head;

        while (curNode != nullptr && curNode->next != nullptr)
        {
            preNode = curNode;
            curNode = curNode->next;

            if (curNode == nullptr)
            {
                break;
            }

            if (curNode->val == preNode->val)
            {
                ListNode *tmpNode = curNode;
                preNode->next = curNode->next;
                delete tmpNode;
                curNode = preNode;
            }
        }

        return head;
    }
};

Python

# LeetCode 83

# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        if head is None:
            return head
        
        curNode = head
        preNode = None
        while curNode is not None and curNode.next is not None:
            preNode = curNode
            curNode = curNode.next

            if curNode is None:
                break

            if curNode.val == preNode.val:
                tmpNode = curNode
                preNode.next = curNode.next
                del tmpNode
                curNode = preNode

        return head