Given a sorted linked list, delete all duplicates such that each element appear only once. 給定一個已排序連結串列，刪除其中重複的元素，使其每個元素只出現一次

Example 1:

Input: 1->1->2
Output: 1->2

Example 2:

Input: 1->1->2->3->3
Output: 1->2->3

Solution：

Python

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

class Solution:
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        dummy = head   #建立一個指向連結串列的指標
        while head and head.next:
            while  head.val == head.next.val:
                if head.next.next:   #如果head.next.next存在，則跨過重複的節點，將head.next.next賦給head.next
                    head.next = head.next.next
                else:   #當head.next.next不存在的時候且head.val == head.next.val，表明最後兩個節點相同，那麼丟棄最後一個節點，跳出迴圈(重要！)
                    head.next = None
                    break
            head = head.next   #如果head與head.next值不同，那麼head向後移動
        return dummy

 

C++思路與Python相同

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) {
        ListNode *dummy = head;
        while(head){
            while (head->next && head->val ==head->next->val){
                if (head->next->next){
                    head->next = head->next->next;
                }
                else{
                    head->next = NULL;
                }
            }
            head = head->next;
        }
        return dummy;     
    }
};