[LeetCode]160. Intersection of Two Linked Lists

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本文介绍了一种高效算法，用于查找两个单链表开始相交的节点。通过巧妙处理不同长度链表的问题，该算法能在O(n)时间内完成任务，并且仅使用O(1)的额外空间。

重点在于两个list的长度。如何可以巧妙地使得两个不同长度的list的指针遍历到同一个位置上去。

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;
        ListNode a, b;
        a = headA;
        b = headB;
        
        while(a!=b){
            if(a==null) a=headB;
            else a=a.next;
            if(b==null) b=headA;
            else b=b.next;
            
        }
        return a;
    }
}