题目:
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.
题意:
给定一个排好序的链表,将重复元素删除。
思路:
只要用两个指针,与当前指针一样则第二个指针继续往后移动,否则第一个指针链接第二个指针,并且移动第一个指针到第二个指针的位置。
以上。
代码如下:
/**
* 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 head;
ListNode* prev = head;
ListNode* curr = head->next;
while(curr != NULL) {
if(curr->val != prev->val) {
prev->next = curr;
prev = curr;
}
curr = curr->next;
}
prev->next = NULL;
return head;
}
};