82. Remove Duplicates from Sorted List II

Leetcode: https://leetcode.com/problems/remove-duplicates-from-sorted-list-ii/

双指针遍历，对比节点值。

• 因为值重复的节点需要全部删掉（不保留distinct），所以对比完成后，多了一个删掉前节点的操作。而且因为 head 节点也可能被删除，需要设置 dummy head 来记录表头。

• 注意随时检查指针的存在性。

• 由于两个指针之间距离是保持固定的，所以也可以用单指针的写法。

双指针

# 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: ListNode) -> ListNode:
        dummy = ListNode()
        dummy.next = head
        if not dummy.next or not dummy.next.next:
            return head
        i, j = dummy, dummy.next.next
        while j:
            if i.next.val == j.val:
                while j and i.next.val == j.val:
                    j = j.next
                    i.next.next = j
                i.next = j
                if not j: break 
            else:
                i = i.next
            j = j.next
        return dummy.next
        

单指针

# 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: ListNode) -> ListNode:
        if not head: return 
        
        dummy = ListNode()
        dummy.next = head
        
        curr = dummy
        while curr.next and curr.next.next:
            if curr.next.val == curr.next.next.val:
                duplicate = curr.next.val
                while curr.next and curr.next.val == duplicate:
                    curr.next = curr.next.next
            else:
                curr = curr.next
        return dummy.next