(1)

class Solution(object):
    def firstUniqChar(self, s):
        """
        :type s: str
        :rtype: int
        """
        letters=‘abcdefghijklmnopqrstuvwxyz‘
        index=[s.index(l) for l in letters if s.count(l) == 1]
        return min(index) if len(index) > 0 else -1

(2)

# Definition for singly-linked list.
 

# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        next=None
        while head:
            head.next,head,next
 
=next,head.next,head
        return next  

(3)

class Solution(object):
    def intersect(self, nums1, nums2):
        """
        :type nums1: List[int]
        :type nums2: List[int]
        :rtype: List[int]
        """
        c1 = collections.Counter(nums1)
        c2 = collections.Counter(nums2)
        
 
return list((c1&c2).elements())#elements返回一個叠代器。元素被重復了多少次，在該叠代器中就包含多少個該元素。元素排列無確定順序，個數小於1的元素不被包含。

leetcode date2018/4/8