给定一个排序链表,删除全部含有重复数字的节点,只保留原始链表中 没有重复出现 的数字。web
输入: 1->2->3->3->4->4->5
输出: 1->2->5svg
输入: 1->1->1->2->3
输出: 2->3spa
利用三个指针head
,t1
,t2
,其关系为head->next = t1
, t1->next = t2
。每次判断t1
和t2
值的状况,并进行标记。若是相等,则标记,并删除t1
节点,更新指针;不等时标记为真,则也须要删除t1
节点,由于以前有和t1
相同的数字,并更新指针;其它则直接更新三个指针。
运行时间0ms,代码以下。指针
/** * Definition for singly-linked list. * struct ListNode { * int val; * struct ListNode *next; * }; */ struct ListNode* deleteDuplicates(struct ListNode* head) { if(head == NULL || head->next == NULL) return head; struct ListNode* front = (struct ListNode*)malloc(sizeof(struct ListNode)); front->next = head; struct ListNode *t1 = head, *t2 = head->next; head = front; bool flag = false; while(t2) { if(t1->val == t2->val) { flag = true; head->next = t2; } else if(flag) { flag = false; head->next = t2; } else head = t1; t1 = t2; t2 = t2->next; } if(flag) { head->next = NULL; } free(front); t1 = front->next; return t1; }