[leetcode] 83. Remove Duplicates from Sorted List @ python

最新推荐文章于 2022-04-06 16:22:53 发布

原创最新推荐文章于 2022-04-06 16:22:53 发布 · 117 阅读

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本文介绍两种删除链表中重复元素的方法：一种是利用集合去重并重新构建链表；另一种是指针法直接在原链表上操作。两种方法分别讨论了时间复杂度和空间复杂度。

原题

Given a sorted linked list, delete all duplicates such that each element appear only once.

Example 1:

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

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

解法1

先遍历链表, 将链表的值转为list, 然后将list转化为集合去掉重复值, 再转化为list升序排列, 最后将list转化为链表
Time: O(n) + O(m), n是链表中节点数量, m是集合的长度
Space: O(1)

代码

# 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
        """
        l = self.getList(head)
        dummy = p = ListNode(0)
        for val in l:
            p.next = ListNode(val)
            p = p.next
        return dummy.next
    
    def getList(self, head):
        l = []
        while head:
            l.append(head.val)
            head = head.next
        return sorted(set(l))

解法2

指针法. 定义cur节点, 如果探测到cur.next节点的值与cur值相同, 则略过cur.next节点
Time: O(n)
Space: O(1)

代码

# 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
        """
        cur = head
        while cur:
            while cur.next and cur.next.val == cur.val:
                # skip cur.next
                cur.next = cur.next.next
            # move to the next node
            cur = cur.next
        return head