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

For example, the following two linked lists: 

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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.

思路

首先要明白题意，如果两个链表相交，这两个链表也不会再分离开。

从而可以先求出链表的长度。如果两个链表相交，对齐它们的尾部，总会在其中一个节点，两链表的值相同。

根据此思路，从而求出链表值相同的那个节点，即可以满足题意。

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# 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
        """
        curA = headA
        curB = headB
        lenA = 0
        lenB = 0
        while (curA):
            curA = curA.next
            lenA += 1
        while (curB):
            curB = curB.next
            lenB += 1
            
        if (curA != curB):
            return None
        else:
            curA = headA
            curB = headB
            if (lenA >= lenB):
                for i in range(lenA-lenB):
                    curA = curA.next
            else:
                for i in range(lenB-lenA):
                    curB = curB.next
            while (curA and curB):
                if (curA == curB):
                    return curA
                else:
                    curA = curA.next
                    curB = curB.next
        return None