Remove Duplicates from Sorted List

算法太简单，就看实现简不简洁，这道题自己写的算法也能过，但是没有答案这么简单，其实说白了就是一个指针在当前位，然后看是否和相邻的相等，如果相等，则把当前节点的next跳到下下个去再来循环，有点类似搭桥的感觉

# 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 and cur.next:
            if cur.val == cur.next.val:
                cur.next = cur.next.next
            else:
                cur = cur.next
        return head

Remove Duplicates from Sorted List II

和上题不同的是，这题一旦发现重复元素则将其全部删去一个不留，自己的代码也能过，但还是逻辑不好看，下面是干净的版本

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        dummy = ListNode('d')
        dummy.next = head
        pre, cur = dummy, head

        while cur:
            while cur.next and cur.val == cur.next.val:
                cur = cur.next
            if pre.next = cur:
                pre = pre.next
            else:
                pre.next = cur.next
            cur = cur.next

        return dummy.next