删除链表中的重复结点

来源：互联网发布：淘宝鹤王阿胶糕是假货编辑：程序博客网时间：2024/04/29 09:49

题目描述
在一个排序的链表中，存在重复的结点，请删除该链表中重复的结点，重复的结点不保留，返回链表头指针。例如，链表1->2->3->3->4->4->5 处理后为 1->2->5

思路
因为相同的结点全部都要删除，我们要设置三个指针，一个指向当前结点，一个指向前一结点，一个指向下一结点，一旦遇到当前结点等于下一结点的情况，那么就一直移动，直到遇到下一个结点不同的状况，再将前一个结点指向此时结点的位置。

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/*
struct ListNode {
    int val;
    struct ListNode *next;
    ListNode(int x) :
        val(x), next(NULL) {
    }
};
*/
classSolution
{
    public:
        ListNode* deleteDuplication(ListNode* pHead)
        {
            if(pHead==nullptr)
                returnpHead;
            ListNode *pPreNode = nullptr;
            ListNode *pNode = pHead;
            while(pNode!=nullptr)
            {
                ListNode *pNextNode = pNode->next;
                bool needDelete = false;
                if(pNextNode!=nullptr && pNode->val==pNextNode->val)
                    needDelete = true;
                if(!needDelete)
                {
                    pPreNode = pNode;
                    pNode = pNode->next;
                }
                else
                {
                    intvalue = pNode->val;
                    ListNode *pToBeDel = pNode;
                    while(pToBeDel!=nullptr && pToBeDel->val == value)
                    {
                        pNextNode = pToBeDel->next;
                        delete pToBeDel;
                        pToBeDel = nullptr;
                        pToBeDel = pNextNode;
                    }
                    if(pPreNode==nullptr)
                        pHead = pNextNode;
                    else
                        pPreNode->next = pNextNode;
                    pNode = pNextNode;
                }
            }
            returnpHead;
        }
};

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