83. Remove Duplicates from Sorted List(Linked List-Easy)

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：对给定的排好序的链表，删除重复的元素，只留下出现一次的元素

思路：当元素和下一个元素比对，如果相同，当前元素的next指针指向下一个元素的next指针。

Language : c

/**
 * Definition for singly-linked list.
 * struct ListNode {
 * int val;
 * struct ListNode *next;
 * };
 */
struct ListNode* deleteDuplicates(struct ListNode* head) { struct ListNode* cur = (int *)malloc(sizeof(struct ListNode)); cur = head; while(cur != NULL){ while(cur->next != NULL && cur->val == cur->next->val){ cur->next = cur->next->next; } cur = cur->next; } return head;
}
  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
 

Language : cpp

/**
 * 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) { ListNode* cur = head; while(cur != NULL){ while(cur->next != NULL && cur->val == cur->next->val){ cur->next = cur->next->next; } cur = cur->next; } return head; }
};
  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
  
19
  
20
  
21
 

Language : python

# Definition for singly-linked list.
# class ListNode(object):
# def __init__(self, x):
# self.val = x
# self.next = None

class Solution(object): def deleteDuplicates(self, head): """ :type head: ListNode :rtype: ListNode """ now = head while head: while head.next and head.val == head.next.val: head.next = head.next.next head = head.next return now
  

 

  
1
  
2
  
3
  
4
  
5
  
6
  
7
  
8
  
9
  
10
  
11
  
12
  
13
  
14
  
15
  
16
  
17
  
18
  
19
 

文章来源: jackcui.blog.csdn.net，作者：Jack-Cui，版权归原作者所有，如需转载，请联系作者。

原文链接：jackcui.blog.csdn.net/article/details/55106317