【Leetcode】83. Remove Duplicates from Sorted List

最新推荐文章于 2024-11-10 15:36:20 发布

原创最新推荐文章于 2024-11-10 15:36:20 发布 · 191 阅读

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#链表 #Leetcode

本文介绍四种不同的算法来解决删除单链表中重复元素的问题，包括使用快慢指针、递归方法、单循环和重复循环的方式。每种方法都有其特点和优势，如递归方法比99.98%的解决方案更快。

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


class Solution1:
    """
    slow&fater pointer
    remember to make slow.next equal to None after jump out of the loop
    """
    def deleteDuplicates(self, head) :
        if head == None: return head
        slow, fast = head, head
        while(fast):
            if fast.val == slow.val:
                fast = fast.next
            else:
                slow.next = fast
                fast = fast.next
                slow = slow.next
        slow.next = None
        return head

class Solution2:
    """
    recursion way use last_val to record last distinct node's val
    faster than 99.98%
    """
    def deleteDuplicates(self, head):
        return self.helper(head, float("inf"))

    def helper(self, head, last_val):
        if head is None: return head
        if head.val == last_val:
            return self.helper(head.next, last_val)
        else:
            head.next = self.helper(head.next, head.val)
            return head

class Solution3:
    """
    use one variable , delete next node if next node's val is the same
    """
    def deleteDuplicates(self, head):
        pNode = head
        while(pNode):
            while(pNode.next and pNode.val == pNode.next.val):
                pNode.next = pNode.next.next
            pNode = pNode.next
        return head

class Solution4:
    """
    use one loop, what is different from solution3 is that, s3 use one loop to get next node
    but s4 use loop again and again to find next node
    """
    def deleteDuplicates(self, head):
        if head is None: return head
        pNode = head
        while(pNode.next):
            if pNode.val == pNode.next.val:
                pNode.next = pNode.next.next
            else:
                pNode = pNode.next
        return head