26.leetcode160_intersection_of_two_linked_lists

1.题目描述

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.

找出双链表交叉处的指针，没有的话返回None。

2.题目分析

题目给的限制条件比较多，比较令人难受的是不能改变链表结构(不能做标记)。所以采用与142题类似的解法，双指针同时前进一个节点找到相同节点，避免n^2的重复比较。

3.解题思路

①从后向前取相同长度(因为相同的是后半截)

②从等长度处开始同时遍历等长度

# 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
        """
        l1=0
        l2=0
        p1=headA
        p2=headB
        while p1!=None:
            l1+=1
            p1=p1.next
        while p2!=None:
            l2+=1
            p2=p2.next
        p1=headA
        p2=headB
        n=l1-l2
        if n>0:
            for i in range(0,n):
                p1=p1.next
        if n<0:
            for i in range(0,-n):
                p2=p2.next
        while p1!=None:
            if p1==p2:
                return p1
            else:
                p1=p1.next
                p2=p2.next
        return None