【55】160. Intersection of Two Linked Lists

160. Intersection of Two Linked Lists

Description Submission Solutions Add to List

• Total Accepted: 112057
• Total Submissions: 372496
• Difficulty: Easy
• Contributors: Admin

 

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.

 

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
//先算长度差，再一起往后走。最后判断下是否为空即可。
    ListNode *getIntersectionNode(ListNode *headA, ListNode *headB) {
        if(!headA || !headA) return NULL;
        ListNode* tmp = headA;
        int lenA = 0, lenB = 0;
        while(tmp){
            tmp = tmp -> next;
            lenA++;
        }
        tmp = headB;
        while(tmp){
            tmp = tmp -> next;
            lenB++;
        }
        if(lenA > lenB){
            int len = lenA - lenB;
            while(len > 0){
                headA = headA -> next;
                len--;
            }
        }else if(lenA < lenB) {
            int len = lenB - lenA;
            while(len > 0){
                headB = headB -> next;
                len--;
            }
        }
        while(headA != headB && headA && headB){
            headA = headA -> next;
            headB = headB -> next;
        }
        
        return headA && headB ? headA : NULL;
    }
};