[LeetCode Road] Intersection of Two Linked Lists - Solution/C++

160. Intersection of Two Linked Lists

Question:

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists: begin to intersect at node c1.

Example:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,6,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node's value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,6,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Input: intersectVal = 2, listA = [1,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node's value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [1,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

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.

Each value on each linked list is in the range [1, 10^9].

Your code should preferably run in O(n) time and use only O(1) memory.

Source code:

Version 1

Idea:
The length of two List may be different, so you should count the length of headA and headB at 17-25 line.
Notice: You must declare the pointer of ListNode as headA & headB.

After counting, you should make those Lists moved to the same position, as below:

(a.) The count of headA is bigger than headB

(b.) The count of headA is smaller than headB

At 39-42 line, it starts comparing each value of List, if find the same reference(address), return headA(or headB).
However, if there is no intersection(can not find the same Listnode), the List of headA or headB points to NULL, hence also return headA(or headB).

/**
 * 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 lenA = 0, lenB = 0;
        ListNode *tempA = headA;
        ListNode *tempB = headB;
        
        while(tempA){
            lenA++;
            tempA = tempA->next;
        }
        
        while(tempB){
            lenB++;
            tempB = tempB->next;
        }
        
        if(tempA!=tempB) return NULL;
        
        if(lenA > lenB){
            for(int i = 0; i < lenA - lenB; i++)
                headA = headA->next;
        }
        
        if(lenB > lenA){
            for(int i = 0; i < lenB - lenA; i++)
                headB = headB->next;
        }
        
        while(headA != headB){
            headA = headA->next;
            headB = headB->next;
        }
        
        return headA;
    }
};

Version 2

Idea:
Let takes Example 2 to explain how efficient method it is:

Go through each node with headA and headB at the same time, meanwhile, check the reference of nodes whether those are the same or not. If one of the List points to NULL(means at the end), restart to go through the other, the illustration is shown in the following,

So, if there is the intersection between two singly linked lists, the method will find. On the contrary, those will point to NULL.

/**
 * 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 *tempA = headA;
        ListNode *tempB = headB;
        while(tempA != tempB){
            tempA = tempA?tempA->next:headB;
            tempB = tempB?tempB->next:headA;
        }
        return tempA;
    }
};