Q160 Intersection of Two Linked Lists

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