LeetCode Intersection of Two Linked Lists

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本文介绍了一种在两个单链表中找到它们开始相交节点的算法。通过分析链表长度差异并调整遍历起点，确保了两链表在相交处同步相遇，实现了O(n)的时间复杂度和O(1)的空间复杂度。

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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.

思路分析：这题考察链表操作，首先要正确理解题意。如果两个链表有intersection，那么最后一个node必然是同一个node，否则就是没有intersection。然后如果我们已经知道两个链表有intersection，如果去找第一个交叉的node呢？最直接的方法是用两个指针同时在两个链表走，但是如何保证它们可以同时到达那个首个交叉的node呢？我们可以让这两个指针的起点不同，保证他们同时走的时候，到达首个交叉的node的距离相等。于是我们想到利用两个链表的长度之差|len1 - len2|.如果len1 > len2,那么就先让pa先走|len1 - len2|步，然后pa和pb一起走，必然同时到达首个交叉的node。对于len1 <= len2的情况，同理可以分析。

AC Code

/**
 * 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) {
        if(headA == null || headB == null) return null;
        ListNode pa = headA;
        ListNode pb = headB;
        int lenA = 1;
        int lenB = 1;
        while(pa.next != null){
            pa = pa.next;
            lenA++;
        }
        while(pb.next != null){
            pb = pb.next;
            lenB++;
        }
        if(pa != pb) return null; //no intersection
        pa = headA;
        pb = headB;
        int diff = Math.abs(lenA - lenB);
        if(lenA > lenB){
            while(diff != 0 ){
                pa = pa.next;
                diff--;
            }
        } else {
            while(diff != 0){
                pb = pb.next;
                diff--;
            }
        }
        while(pa != pb){
            pa = pa.next;
            pb = pb.next;
        }
        return pa;
    }
}