Leetcode 82

Question

Solution

首先创建两个指针：pre, cur。pre指向空，cur指向head。如果出现多个相同元素，cur向后移动直到指向新的元素，pre.next指向cur。如果元素不重叠，pre移到cur的位置，cur向后移动一位。

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

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        ans = pre = ListNode(0)
        ans.next = head
        cur = head
        while cur:
            if cur.next and cur.val == cur.next.val:
                temp = cur.val
                while cur and cur.val == temp:
                    cur = cur.next
                pre.next = cur
            else:
                pre = cur
                cur = cur.next
        return ans.next