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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.
分析：将前一个数与后一个数比较，如果不同且出现次数为1，则添加到结果里面。特别要注意最后一个结点：如果与前一个结点相同，则不必添加；如果与前一个结点不同，则必须要添加。运行时间14ms。思考：把line30~32为啥会出现wrong answer？
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 || !head->next) return head; 13 14 ListNode *result = new ListNode(0); 15 ListNode *preHead = result; 16 int times = 1; 17 while(head->next){ 18 if(head->val == head->next->val) times++; 19 else{ 20 if(times == 1){ 21 result->next = head; 22 result = result->next; 23 } 24 times = 1; 25 } 26 head = head->next; 27 } 28 if(times == 1){ 29 result->next = head; 30 result = result->next; 31 } 32 result->next = NULL; 33 34 return preHead->next; 35 } 36 };
Remove Duplicates from Sorted List II
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原文地址：http://www.cnblogs.com/amazingzoe/p/4463591.html
