lc 相交链表

链接：https://leetcode-cn.com/problems/intersection-of-two-linked-lists/

代码：

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

class Solution(object):
    def getIntersectionNode(self, headA, headB):
        """
        :type head1, head1: ListNode
        :rtype: ListNode
        """
        if headA is None or headB is None:
            return None
        curA = headA
        print type(headA)
        curB = headB
        len_a = 0
        len_b = 0
        while curA is not None:
            len_a += 1
            curA = curA.next
        while curB is not None:
            len_b += 1
            curB = curB.next
        curA = headA
        curB = headB
        
        minus = len_a - len_b
        print minus
        if minus >= 0:
            while curA is not None and minus > 0:
                curA = curA.next
                minus -= 1
        else:
            while curB is not None and minus < 0:
                curB = curB.next
                minus += 1
        while curA is not None and curB is not None and curA != curB:
            curA = curA.next
            curB = curB.next
        if curA is None:
            return None
        else:
            return curA

思路：短的链表先走长链表与短链表之差，然后再一起走，返回交点