160. Intersection of Two Linked Lists 要再做一遍

寻找链表交点

最新推荐文章于 2022-08-06 10:20:24 发布

原创最新推荐文章于 2022-08-06 10:20:24 发布 · 197 阅读

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本文介绍了一种在不改变链表结构的前提下找到两个单链表相交起始节点的方法，并提供了一个运行时间为O(n)，空间复杂度为O(1)的解决方案。

/**
 * 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) {
        int lenA = length(headA);
        int lenB = length(headB);
        
        while(lenA>lenB){
            headA = headA.next;
            lenA--;
        }
        while(lenA<lenB){
            headB = headB.next;
            lenB--;
        }
        while(headA!=headB){
            headA = headA.next;
            headB = headB.next;
        }
        return headA;
    }
    
    private int length(ListNode node){
        int len=0;
        while(node!=null){
            node = node.next;
            len++;
        }
        return len;
    }
}

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.

开始的思路是：想着把list反过来再比较，然后发现不可以改变链表的结构所以通过不了test

然后这个思路是，先算出list的length，然后根据谁大谁小，把两个的指针移到同一个地方（短的那一个list的开头）开始进行比较

相同就返回，一直没有相同的那么最后headA会成为null。