这篇文章主要讲解了“怎么用python删除重复的元素”,文中的讲解内容简单清晰,易于学习与理解,下面请大家跟着小编的思路慢慢深入,一起来研究和学习“怎么用python删除重复的元素”吧!
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; }
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; } };
Language : python
# Definition for singly-linked list.# class ListNode(object):# def __init__(self, x):# self.val = x# self.next = Noneclass Solution(object):def deleteDuplicates(self, head):""" :type head: ListNode :rtype: ListNode """ now = headwhile head:while head.next and head.val == head.next.val: head.next = head.next.next head = head.nextreturn now
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原创文章,作者:kirin,如若转载,请注明出处:https://blog.ytso.com/tech/opensource/220254.html