LeetCode – Refresh – Intersection of Two Linked Lists

Find the common length part, then check with two pointers.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (!headA || !headB) return NULL;
        int la = 1, lb = 1;
        ListNode *runner = headA;
        while (runner->next) {
            runner = runner->next;
            la++;
        }
        runner = headB;
        while (runner->next) {
            runner = runner->next;
            lb++;
        }
        if (la > lb) {
            while (la > lb) {
                headA = headA->next;
                la--;
            }
        } else {
            while (lb > la) {
                headB = headB->next;
                lb--;
            }
        }
        while (headA && headA->val != headB->val) {
            headA = headA->next;
            headB = headB->next;
        }
        return headA;
    }
};