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.
Notes:
- If The linked lists has no intersection at all, return
null .
- The linked lists must retain their original structure after the function returns.
- You may assume there is no cycles anywhere in the entire linked structure.
- Your code should preferably run in O (n) time and use only O (1) memory.
The problem is still to understand the topic to ask what. Here's the intersection linked list by looking at the example we can see that its length can be different, but from the beginning of the Intersect to the end of the length is the same, so that the extra is only a few of the previous values, And there must be a longer length of that linked list the value of the extra section.
So we first calculate the length of the two list, and give the value of the extra part of the longer that waive off. So the two list of the same length is very easy to find intersect place.
The code is as follows. ~
/** * Definition for singly-linked list. * public class ListNode {* int val; * ListNode Next; * ListNode (int x) {* val = x; * Next = Null *} *} */public class Solution {public ListNode Getintersectionnode (ListNode heada, ListNode headb) {int lengtha=0; int lengthb=0; ListNode Curr1=heada; ListNode curr2=headb; while (curr1!=null) {lengtha++; Curr1=curr1.next; } while (Curr2!=null) {lengthb++; Curr2=curr2.next; } Curr1=heada; CURR2=HEADB; if (LENGTHA>LENGTHB) {for (int i=0;i<lengtha-lengthb;i++) {curr1=curr1.next; }}else{for (int i=0;i<lengthb-lengtha;i++) {curr2=curr2.next; }} while (CURR1!=CURR2) {curr1=curr1.next; Curr2=curr2.next; } return CURR1; }}
[Leetcode] intersection of Linked Lists