Intersection of Two Linked Lists

最新推荐文章于 2024-06-24 11:25:34 发布

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本文介绍了一种高效算法，用于在给定的两个单链表中找到它们可能的相交节点。通过避免使用额外的空间，并确保算法在O(n)时间内运行且仅使用O(1)内存，该算法提供了解决此类问题的有效解决方案。

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 ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if( headA==null || headB==null) return null;
        int lenA = getLen(headA);
        int lenB = getLen(headB);
        ListNode pa = headA;
        ListNode pb = headB;
        int dis = Math.abs(lenA - lenB);
        if(lenA > lenB){
            for(int i = 0; i < dis; i++){
                pa = pa.next;
            }
        } else if(lenB > lenA){
            for(int i = 0; i < dis; i++){
                pb = pb.next;
            }
        }
        while(pa != null && pa.val != pb.val){
            pa = pa.next;
            pb = pb.next;
        }
        return pa;
    }
    public int getLen(ListNode head){
        int len = 0;
        for(ListNode n = head; n != null; n = n.next){
            len ++;
        }
        return len;
    }
}

思路：

1、用栈。将两条链表存入栈中，然后逐个pop比较。需消耗较多空间，不能满足题目要求。

2、快行指针思想，遍历时，沿较长的链表先走。