160. 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.

Solution 1: firstly get the sizes of the two linked lists A and B, then calculate the difference of the sizes and move the head of the longer list forward with the difference value, then compare the elements of two lists one by one till finding the same elements.

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    int getSize(ListNode *head){
        int size=0;
        while (head){
            size++;
            head=head->next;
        }
        return size;
    }
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        int sizeA=getSize(headA), sizeB=getSize(headB);
        if (sizeA==0 || sizeB==0) return NULL;
        int sizeDiff=sizeA-sizeB;
        if (sizeDiff>=0){
            while (sizeDiff>0){
                headA=headA->next;
                sizeDiff--;
            }
        }
        else {
            while(sizeDiff<0){
                headB=headB->next;
                sizeDiff++;
            }
        }
        while (headA!=NULL){
            if (headA->val==headB->val){
                return headA;
            }
            else {
                headA=headA->next;
                headB=headB->next;
            }
        }
        return NULL;
    }
};

Solution 2:  use the idea of cycles, tranverse each list at the same time, when one list is tranversed to the end we jump to the head of the other list and continue the tranverse. The two pointers will be equal finally with two possible conditions: a) intersect at the node b) don't have the intersection, the two pointers will be NULL. The reason for the equal pointers is that the tranverse length of the two pointers are the same, which is the sum of the two lists' length. (so smart....)

class Solution {
public:
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if (!headA || !headB) return NULL;
        ListNode *a = headA, *b = headB;
        while (a != b) {
            a = a ? a->next : headB;
            b = b ? b->next : headA;
        }
        return a;
    }
};