# Leetcode "intersection of the Linked Lists" Python implementation

Source: Internet
Author: User

title :

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

For example, the following, linked lists:

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

Begin to intersect at node C1.

code : OJ online test via runtime:1604 ms

`1 #Definition for singly-linked list.2 #class ListNode:3 #def __init__ (self, x):4 #self.val = x5 #Self.next = None6 7 classSolution:8     #@param listnodes9     #@return The intersected ListNodeTen     defGetlistlen (self,head): OneLength =0 A          whileHead is  notNone: -Length + = 1 -Head =Head.next the         returnlength -              -     defGetintersectionnode (self, Heada, headb): -         ifHeada isNoneorHeadb isNone: +             returnNone -          +HA =Heada AHB =headb at          -LenA =Self.getlistlen (HA) -LenB =Self.getlistlen (HB) -          -         ifLenA >LenB: -Distance = LenA-LenB in              forIinchRange (0,distance): -HA =Ha.next to         ifLenA <LenB: +Distance = LenB-LenA -              forIinchRange (0,distance): theHB =Hb.next *          \$intersection =NonePanax Notoginseng          whileHA is  notNone andHb is  notNone: -             ifHA = =HB: the                 returnHA +             Else: AHA =Ha.next theHB =Hb.next +         returnIntersection`

Ideas :

1. First record the length of the two linked list

2. Double pointers to two lists point to the long list first step, get a new linked list table header

3. Compare the values of each pointer of the two list one after the other equal: until the equality is found, or to none

Leetcode "intersection of the Linked Lists" Python implementation

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