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.

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
    
    public int getListLength(ListNode N){
        int n = 0;
        ListNode p = N;
        while(p != null){
            n++;
            p = p.next;
        }
        return n;
    }
    
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
      
        ListNode intersect = null;
        ListNode pA = headA;
        ListNode pB = headB;
        int A = getListLength(headA);
        int B = getListLength(headB);
        if(A >= B){
            int m = A-B;
            while(m != 0){
                pA = pA.next;
                m--;
            }
        }
        else{
            int m = B-A;
            while(m != 0){
                pB = pB.next;
                m--;
            }
        }
        
        while(pA != null){
            if(pA == pB){
                intersect = pA;
                break;
            }
            else{
                pA = pA.next;
                pB = pB.next;
            }
        }
        
        return intersect;
    }
}

思路很简单，长的链表先移动指针至和短的链表“同一起跑线”，然后同时移动指针并比较是否相同。