letecode [83] -
作者:互联网
Given a sorted linked list, delete all duplicates such that each element appear only once.
Example 1:
Input: 1->1->2 Output: 1->2
Example 2:
Input: 1->1->2->3->3 Output: 1->2->3
题目大意:
给定排序链表,删除其中重复的元素。
理 解 :
遍历链表,cur指向当前节点,若它后面的节点A的值与cur的值相等,则删除它后面的节点A,即cur->next指向A后面的节点B;
若不等,则cur后移一个节点,即cur = cur->next;
代 码 C++:
/** * Definition for singly-linked list. * struct ListNode { * int val; * ListNode *next; * ListNode(int x) : val(x), next(NULL) {} * }; */ class Solution { public: ListNode* deleteDuplicates(ListNode* head) { if(head==NULL) return NULL; int val = head->val; ListNode* cur = head; while(cur->next!=NULL){ if(cur->next->val==val){ cur->next = cur->next->next; }else{ cur = cur->next; val = cur->val; } } return head; } };
运行结果:
执行用时 : 16 ms 内存消耗 : 9.5 MB
标签:head,NULL,ListNode,cur,val,next,83,letecode 来源: https://www.cnblogs.com/lpomeloz/p/10984318.html