【LeetCode】83. Remove Duplicates from Sorted List 解题报告（Python）

题目分析：

这一题是给定一个排序链表，删除所有重复的元素，使得每个元素只出现一次。如:
输入: 1->1->2->3->3
输出: 1->2->3

解题思路：（描述的稀里糊涂，希望你能看懂）

每次检查当前后面一位值是够相同，相同就事当前位的下一位为下下一位。

测试代码：（非提交代码，提交代码在下面）

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

def print_listNode(listNode: ListNode):
    if not listNode:
        print(None)
        return
    bf_listNode = listNode
    while bf_listNode.next:
        print(bf_listNode.val, end='-->')
        bf_listNode = bf_listNode.next
    print(bf_listNode.val)

def creat_listNode(nums: list):
    if not nums: return None
    bf_head = head = ListNode(nums[0])
    for i in range(1, len(nums)):
        head.next = ListNode(nums[i])
        head = head.next
    head.next = None
    return bf_head

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        dummy = head
        while dummy:
            while dummy.next and dummy.val == dummy.next.val:
                dummy.next = dummy.next.next
            dummy = dummy.next
        return head

print_listNode(Solution().deleteDuplicates(creat_listNode([1,1,2])))
print_listNode(Solution().deleteDuplicates(creat_listNode([1,1,2,3,3])))
print_listNode(Solution().deleteDuplicates(creat_listNode([0,0,0,0,3])))

提交代码：（Runtime: 48 ms, faster than 82.80%）

class Solution:
    def deleteDuplicates(self, head: ListNode) -> ListNode:
        dummy = head
        while dummy:
            while dummy.next and dummy.val == dummy.next.val:
                dummy.next = dummy.next.next
            dummy = dummy.next
        return head