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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.
这里对于链表去重是要将有重复的链表节点全部都去掉，一个都不能留，思路还是比较简单的，跟前面的I那一题实际上差不多，代码如下所示：
1 /** 2 * Definition for singly-linked list. 3 * struct ListNode { 4 * int val; 5 * ListNode *next; 6 * ListNode(int x) : val(x), next(NULL) {} 7 * }; 8 */ 9 class Solution { 10 public: 11 ListNode* deleteDuplicates(ListNode* head) { 12 if(!head) return NULL; 13 ListNode * helper = new ListNode(INT_MAX); 14 helper->next = head; 15 ListNode * prev = helper; 16 ListNode * curr = head; 17 while(curr){ 18 if(curr->next && curr->val == curr->next->val){ 19 ListNode * tmpNode = curr->next; 20 while(tmpNode->next && curr->val == tmpNode->next->val){ 21 tmpNode = tmpNode->next; 22 } 23 prev->next = tmpNode->next; 24 curr = prev->next; 25 }else{ 26 prev = curr; 27 curr = curr->next; 28 } 29 } 30 return helper->next; 31 } 32 };
LeetCode OJ：Remove Duplicates from Sorted List II（链表去重II）
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原文地址：http://www.cnblogs.com/-wang-cheng/p/5005052.html
