Given a sorted linked list, delete all duplicates such that each element appear only once.
For example,
Given 1->1->2
, return 1->2
.
Given 1->1->2->3->3
, return 1->2->3
.
思路:两个指针,一个指向当前节点cur,一个指向下一个节点nx,终止条件,也就是比较了len-1次。
比较若不等,则两指针均后移,若相等则删除节点,且cur指针不变,nx指针后移。
代码:
class Solution { public: int getLength(ListNode *head){ int length=0; while(head){ ++length; head=head->next; } return length; } ListNode *deleteDuplicates(ListNode *head) { if(head==NULL) return NULL; int len=getLength(head); int i=1; ListNode* res=head; ListNode* cur=res; ListNode* nx=res->next; while(i<len){ if(cur->val==nx->val){ cur->next=cur->next->next; nx=nx->next; }else{ cur=cur->next; nx=nx->next; } ++i; } return res; } };