leetcode--Intersection of Two Linked Lists

最新推荐文章于 2020-12-21 00:46:23 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。该算法运行时间为O(n)，且只使用常数级别的额外内存。通过比较两个链表的长度，确定起始链表并进行适当调整，最终找到相交节点。

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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.

/**
 * 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 h1 = headA;  
        ListNode h2 = headB;  
        int count1 = 1, count2 = 1;  
        while(h1.next != null){  
            count1++;  
            h1 = h1.next;  
        }  
        while(h2.next != null){  
            count2++;  
            h2 = h2.next;  
        }  
        if(h1 != h2) return null;  
        else{  
            int count = Math.abs(count1 - count2);  
            if(count2 > count1){  
                h1 = headB;  
                h2 = headA;   
            }  
            else{  
                h1 = headA;  
                h2 = headB;  
            }  
            while((count--) > 0){  
                h1 = h1.next;  
            }  
            while(h1 != null&&h2 != null && h1 != h2){  
                h1 = h1.next;  
                h2 = h2.next;  
            }  
            return h1;  
        }  
    }
}