Intersection of Two Linked Lists

两个链表的交集

题目

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.

解析重点

1.我们注意到两个链表如果有交集，那么相交之后的的node都是相交的，因此，我们只需要找出相交的node即可
2.如果两个链表长度相同，我们只需要一起迭代即可，但是两个链表长度不同呢，怎么办？可以用A+B=B+A的形式，
即两个链表一起迭代，a迭代完后迭代b，同时b迭代完后a，第二轮就可以知道相交的点了。

java代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) {
 *         val = x;
 *         next = null;
 *     }
 * }
 */
public class Solution {
   public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
    //boundary check
    if(headA == null || headB == null) return null;
    
    ListNode a = headA;
    ListNode b = headB;
    
    //if a & b have different len, then we will stop the loop after second iteration
    while( a != b){
    	//for the end of first iteration, we just reset the pointer to the head of another linkedlist
        a = a == null? headB : a.next;
        b = b == null? headA : b.next;    
    }
    
    return a;
}
}