Remove Duplicates from Sorted List II

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Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

For example,
Given 1->2->3->3->4->4->5, return 1->2->5.

Given 1->1->1->2->3, return 2->3.

解题思路:新建一个链表，遍历原来链表，若不重复则将结点插入新链表，重复则跳过.为了避免遍历最后为重复结点，最后需将新链表末尾置NULL

#include<iostream>
#include<vector>
using namespace std;
//Definition for singly - linked list.
struct ListNode {
	int val;
	ListNode *next;
	ListNode(int x) : val(x), next(NULL) {}
};
ListNode *deleteDuplicates(ListNode *head) {
	ListNode*ResultList = new ListNode(0);
	ListNode*TmpResultNode = ResultList;
	ListNode*NorepeatNode = head;
	while (NorepeatNode != NULL){
		ListNode*PreNode = NorepeatNode;
		while (NorepeatNode->next != NULL&&NorepeatNode->val == NorepeatNode->next->val)
			NorepeatNode = NorepeatNode->next;
		if (NorepeatNode == PreNode)  //如果不重复，插入
		{
			TmpResultNode->next = NorepeatNode;
			TmpResultNode = TmpResultNode->next;
		}
		NorepeatNode  = NorepeatNode  -> next;
	}
	TmpResultNode->next = NULL;       //尾部置NULL
	return ResultList->next;
}

Remove Duplicates from Sorted List II

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原文地址：http://blog.csdn.net/li_chihang/article/details/43372147