82. &83 Remove Duplicates from S

题目要求：

Given a sorted linked list, delete all nodes that have duplicate numbers, leaving only distinct numbers from the original list.

Eexamples：

Given 1->2->3->3->4->4->5, return 1->2->5.
Given 1->1->1->2->3, return 2->3.

解题思路：

• 指针移动示意图

• 需要遍历链表所有的元素，同时保证当前元素的后一个元素不为空，为空肯定就不会有元素和当前元素相同。

• 只要本元素和前一个元素相同就指针后移。

• else语句开始的链表相连操作需要注意！！！

代码：

class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
    
    def __repr__(self):
        if self is None:
            return "Nil"
        else:
            return "{} -> {}".format(self.val, repr(self.next))

class Solution(object):
    def deleteDuplicates(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        dummy = ListNode(0)
        pre, cur = dummy, head
        while cur:
            if cur.next and cur.next.val == cur.val:
                val = cur.val;
                while cur and cur.val == val:
                    cur = cur.next
                pre.next = cur
            else:
                pre.next = cur
                pre = cur
                cur = cur.next
        return dummy.next

if __name__ == "__main__":
    head, head.next, head.next.next = ListNode(1), ListNode(2), ListNode(3)
    head.next.next.next, head.next.next.next.next = ListNode(3), ListNode(4)
    head.next.next.next.next.next, head.next.next.next.next.next.next = ListNode(4), ListNode(5)
    print(Solution().deleteDuplicates(head))