Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

A:          a1 → a2
                   ↘
                     c1 → c2 → c3
                   ↗            
B:     b1 → b2 → b3

begin to intersect at node c1.

Notes:

• If the two linked lists have no intersection at all, return null.
• The linked lists must retain their original structure after the function returns.
• You may assume there are no cycles anywhere in the entire linked structure.
• Your code should preferably run in O(n) time and use only O(1) memory.

解题思路：

此题需要注意的两点是返回后，需要保持原来的链表结构，同时对时间复杂度的要求为 O(n)，空间负责度的要求为 O(1)。

由于空间复杂度的限制，故排除用Map保存一个链表结点，然后与另一个链表比对的方法。

观察到如果两个链表有相同的交集，那么它们从相同的结点开始，后面长度都是相同的。因此，可以先计算两个链表的长度，然后比较长度的差值 diff，较长的那个先走 diff 步，然后和较短的那个同时遍历。如果指针指向同一元素，则为交点，否则 A 和 B 没有交点。

Python实现：

# 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
        """
        A = A1 = headA; B = B1 = headB
        lenA = lenB = 0
        # 计算链表 A,B 的长度
        while A != None:
            lenA += 1
            A = A.next
        while B != None:
            lenB += 1
            B = B.next
        # 尾对齐链表
        if lenA > lenB:
            diff = lenA - lenB
            while diff > 0:
                A1 = A1.next
                diff -= 1
        else:
            diff = lenB - lenA
            while diff > 0:
                B1 = B1.next
                diff -= 1
        # 从对应的位置开始比较
        while A1 != None and B1 != None and A1.val != B1.val:
            A1 = A1.next
            B1 = B1.next
        return A1  # 或者返回B1