leetcode链表--10、remove-duplicates-from-sorted-list（删除排好序链表中重复结点）

题目描述

 

Given a sorted linked list, delete all duplicates such that each element appear only once.

For example,Given1->1->2, return1->2.

Given1->1->2->3->3, return1->2->3.

 

解题思路：

定义两个指针pre、pNode

其中pNode负责向后查找下一连接在要返回链表的结点，其中pre负责每次记录某一个值得第一个结点。

如果pNode->val == pre->val，pNode就向下一结点移动，直至不相等pre->next = pNode

需要注意的是：如果pNode->next == NULL即pNode为最后一个节点了，且值还相等则pre->next = NULL直接返回头指针

#include <iostream>
#include <malloc.h>
using namespace std;
struct ListNode {
    int val;
     ListNode *next;
     ListNode(int x) : val(x), next(NULL) {}
 };
ListNode *CreateList(int n)
{
    ListNode *head;
    ListNode *p,*pre;
    int i;
    head=(ListNode *)malloc(sizeof(ListNode));
    head->next=NULL;
    pre=head;
    for(i=1;i<=n;i++)
    {
        p=(ListNode *)malloc(sizeof(ListNode));
        cin>>p->val;
        pre->next=p;
        pre=p;
    }
    p->next=NULL;

    return head->next;
}
/*-------------------------输出链表-----------------------------------*/
void PrintList(ListNode *h)
{
    ListNode *p;

    p=h;//不带空的头结点
    while(p)
    {
        cout<<p->val<<" ";
        p=p->next;
        cout<<endl;
    }
}
class Solution {
public:
    ListNode *deleteDuplicates(ListNode *head) {
        if(head == NULL)
            return NULL;
        ListNode *pre = head;
        ListNode *pNode = head;
        while(pNode != NULL)
        {
            while(pNode->val == pre->val)
            {
                if(pNode->next != NULL)
                    pNode = pNode->next;
                else
                {
                    pre->next = NULL;
                    return head;
                }
            }
            pre->next = pNode;
            pre = pre->next;
        }
        return head;
    }
};
int main()
{
    int n1;
    ListNode *h1;
    cout<<"输入链表1的结点数目"<<endl;
    cin>>n1;
    h1 = CreateList(n1);
    cout<<"链表1为："<<endl;
    PrintList(h1);
    Solution s;
    ListNode *h2;
    h2 = s.deleteDuplicates(h1);
    cout<<"去重后链表为："<<endl;
    PrintList(h2);
    return 0;
}