Intersection of Two Linked Lists

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.

Notes:

• 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.

Runtime: 52ms. 

/**
 * 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;
        if(!headA || !headB) return NULL;
        
        // compute the length of A and B
        ListNode* moveA = headA;
        int lengthA = 1;
        while(moveA->next) {
            moveA = moveA->next;
            lengthA++;
        }
        
        ListNode* moveB = headB;
        int lengthB = 1;
        while(moveB->next) {
            moveB = moveB->next;
            lengthB++;
        }
        
        if(moveA->val != moveB->val) return NULL;
        
        // ensure A B are of same length
        moveA = headA;
        moveB = headB;
        while(lengthA > lengthB) {
            moveA = moveA->next;
            lengthA--;
        }
        while(lengthB > lengthA) {
            moveB = moveB->next;
            lengthB--;
        }
        
        // get the intersection
        ListNode* result = moveA;
        while(moveB) {
            if(moveA->val != moveB->val)
                result = moveA->next;
            moveA = moveA->next;
            moveB = moveB->next;
        }
        return result;
    }
};

Analyse: A:  --------x------

　　　　　　　　　　　　　　-------c-----

　　　　    B:　-----y-------

x + c + y = y + c + x

Runtime: 52ms. 

/**
 * 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) {
        ListNode* moveA = headA, *moveB = headB;
        
        while(moveA != moveB) {
            moveA = moveA ? moveA->next : headB;
            moveB = moveB ? moveB->next : headA;
        }
        return moveA;
    }
};