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Description of the problem:

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.



  • 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.


Special thanks to @stellari for adding this problem and creating all test cases.


The idea of solving the problem:

If the two LinkedList intersect, the latter part is the same.

This shows that we can begin to search from the place where we are boarding.

Move the pointer of the long list to the same starting point as the short list length, and start comparing whether it is the same.



 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
class Solution {
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(!headA || !headB)
            return NULL;
        int lenA = getLen(headA);
        int lenB = getLen(headB);
        ListNode* curA = headA;
        ListNode* curB = headB;
        if(lenA > lenB){
            int diff = lenA - lenB;
            while(diff > 0){
                curA = curA->next;
        }else if(lenA < lenB){
            int diff = lenB - lenA;
            while(diff > 0){
                curB = curB->next;
        while(curA && curB){
            if(curA == curB)
                return curA;
            curA = curA->next;
            curB = curB->next;
        return NULL;
    int getLen(ListNode *head){
        int ret = 0;
        ListNode *cur = head;
            cur = cur->next;
        return ret;


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