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废话不多说开始今天的解题Day~

83. Remove Duplicates from Sorted List

Question

Given the head of a sorted linked list, delete all duplicates such that each element appears only once. Return the linked list sorted as well.

Example

Example1

Input: head = [1,1,2]
Output: [1,2]

Example2

Input: head = [1,1,2,3,3]
Output: [1,2,3]

Constraints

• The number of nodes in the list is in the range [0, 300].
• -100 <= Node.val <= 100
• The list is guaranteed to be sorted in ascending order.

解题

题目

首先先简单的翻译一下题目
给一个排序过後的linked list，要把其中重复的节点删除。

Think

作法大致上是这样

• once用来存有出现一次的值，再抓到每个节点的值的时候就进来判断有没有遇到过，没有就存入。
• 有出现过的话，就把前一个节点的next指向现在这个节点的next，就完成啦~
• C版本的话，只差在判断是否重复的方式不是用阵列去存，因为後来发现因为有排序过，只要存一个判断有没有重复就好~

Code

Python

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution:
    def deleteDuplicates(self, head: Optional[ListNode]) -> Optional[ListNode]:
        once = []
        start = head
        previous = None
        
        while start != None:
            print(start.val)
            if start.val not in once:
                once.append(start.val)
                previous = start
            else:
                previous.next = start.next

            start = start.next
            
        return head

C

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     struct ListNode *next;
 * };
 */


struct ListNode* deleteDuplicates(struct ListNode* head){
    int num = -101;
    struct ListNode* start = head;
    struct ListNode* previous = 0;
    
    while (start != 0){
        // printf("%d\n", start->val);
        // printf("%d\n", start->next);
        
        if (start == head){
            num = start->val;
            previous = start;
        } else {
            if (start->val > num){
                num = start->val;
                previous = start;
            } else if (start->val == num){
                previous->next = start->next;
            }
        }
        
        start = start->next;
    }
    return head;
}

Result

• Python

  • 

• C

  • 

大家明天见