Title description:
Method one (violence method):
/**
* 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) {
if(headA == null || headB == null){
return null;
}
ListNode q,p;
for(q = headA;q != null;q = q.next){
for(p = headB;p != null;p = p.next){
if(q == p){
return p;
}
}
}
return null;
}
}
Method two (dual pointer):
Suppose that linked list A and linked list B intersect at a certain point, p points to the head of A, and q points to the head of B. First of all, long-linked list + short-linked list = short-linked list + long-linked list, combined with the same distance after the last intersecting node, we can know that as long as p traverses A, traverse B, q after traversing B, traverse A, these two The two pointers will start to meet at the intersection node, that is, the beginning of the intersection node is found.
/**
* 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) {
if(headA == null || headB == null){
return null;
}
ListNode p = headA,q = headB;
while(p != q){
p = p == null ? headB : p.next;
q = q == null ? headA : q.next;
}
return p;
}
}
result:
