删除链表中重复的结点(C++)

本来挺容易的一道题，做起来就有很多问题，解决了之后，也感觉到自己的一些进步

首先要排除一些好解决的问题，链表结点就容易出现一些问题

最重要的就是当headPtr节点为NULL了，这时候就退出了，但是headBefore的值可能还要放入ret链表中去

在最后headPtr和headBefore两个节点的值比较的过程之中，分两个的值相等与不相等，进行后续处理得出最后的结果

首先ret：指向返回链表首结点

retPtr：指向返回链表的末尾节点，用来增长链表

pHeadPtr:指向pHead链表，用来移动判断

pHeadBefore：指向pHead链表，用来与pHeadPtr判断出来不想等的值，就把这个值产生新的节点加入ret

总体思路就是

1、先讨论链表为NULL的和链表只有一个结点的，将这两种情况排除之后，就是判断值pHeadBefore->value和pHeadPtr->value是否相等

2、不相等，就加入pHeadBefore->value；相等就移动pHeadPtr，直到不相等，使得pHeadBefore=pHeadPtr,pHeadPtr=pHeadPtr->next

代码如下：

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#include<iostream> using namespace std;   struct ListNode {     int val;     struct ListNode *next;     ListNode(int x) :         val(x), next(NULL) {     } }; class Solution { public:     ListNode* deleteDuplication(ListNode* pHead)     {         if (pHead == NULL)         {             return NULL;         }         //最后的返回链表         ListNode* ret = NULL;         //         ListNode* retPtr = NULL;         //现链表的前一个节点         ListNode* pHeadBefore = pHead;         //指向现节点         ListNode* pHeadPtr = pHead->next;         //retPtr = ret;         while (pHeadPtr != NULL)         {             if (pHeadPtr->val != pHeadBefore->val)             {                 if (ret == NULL)                 {                     ret = new ListNode(pHeadBefore->val);                     retPtr = ret;                 } else {                     retPtr->next = new ListNode(pHeadBefore->val);                     retPtr = retPtr->next;                 }                 pHeadBefore = pHeadPtr;                 pHeadPtr = pHeadPtr->next;                 if (pHeadPtr == NULL)                 {                     retPtr->next= new ListNode(pHeadBefore->val);                 }             } else {                 while(pHeadPtr != NULL && pHeadPtr->val == pHeadBefore->val)                 {                     pHeadPtr = pHeadPtr->next;                 }                 if (pHeadPtr != NULL)                 {                     pHeadBefore = pHeadPtr;                     pHeadPtr = pHeadPtr->next;                       if (pHeadPtr == NULL)                     {                         if (ret != NULL)                         {                             retPtr->next = new ListNode(pHeadBefore->val);                         }                     }                 }                   }         }         if (ret == NULL)         {             //只有一个节点的情况             if (pHeadBefore->next == NULL)             {                 return new ListNode(pHeadBefore->val);             }           }         return ret;     } }; void printListNode(ListNode* root) {     if (root == NULL)     {         cout<<"NULL"<<endl;     }     while (root != NULL)     {         cout<<root->val<<" ";         root = root->next;     }     cout<<endl; } int main() {     ListNode l1(1);     ListNode l2(2);     ListNode l3(3);     ListNode l4(3);     ListNode l5(4);     ListNode l6(4);     ListNode l7(5);     l1.next = &l2;     l2.next = &l3;     l3.next = &l4;     l4.next = &l5;     l5.next = &l6;     l6.next = &l7;     printListNode(&l1);     printListNode(Solution().deleteDuplication(&l1));     return 0; }