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  • leetcode 83. Remove Duplicates from Sorted List

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

    For example,
    Given 1->1->2, return 1->2.
    Given 1->1->2->3->3, return 1->2->3.

    # 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
            """
            node = head
            while node:
                tmp = node
                val = node.val            
                while node.next and node.next.val == val:
                    node = node.next
                # assert node.next is None or node.next.val != val
                tmp.next = node.next
                node = node.next
            return head
    
    # 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
            """
            dummy = cur = ListNode(None)
            node = head
            while node:
                if node.val != cur.val:
                    cur.next = node
                    cur = cur.next
                node = node.next
            cur.next = None
            return dummy.next
    

    发现其他人的解法和我的都不一样,每次上一个节点比较,如果数值相同,则直接删除当前节点。

    # 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
            """
            if not head: return None      
            node = head
            while node.next:
                if node.next.val == node.val:
                    node.next = node.next.next
                else:
                    node = node.next
            return head
    

     递归解法:

    # 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
            """
            if not head: return None              
            if not head.next: return head        
            head.next = self.deleteDuplicates(head.next)
            if head.val == head.next.val:
                return head.next
            else:
                return head
    
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  • 原文地址:https://www.cnblogs.com/bonelee/p/8878551.html
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