【Leetcode】Intersection of Two Linked Lists_it 21.4 linked 12:1349 may-优快云博客

本文介绍了一种寻找两个单链表开始相交节点的方法，并提供了两种实现方式：一种利用链表长度差，另一种通过节点值累积求和。两种方法均在O(n)时间内完成，并使用O(1)额外空间。

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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.

Java:

第一个是用长度差 O(2(m+n))

/**
 * 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=length(headA);
        int lenB=length(headB);
        int diff=Math.abs(lenA-lenB);
        while(diff>0){
            if(lenA>lenB) headA=headA.next;
            else headB=headB.next;
            diff--;
        }
        while(headA!=null&&headB!=null){
            if(headA.val==headB.val) return headA;
            headA=headA.next;
            headB=headB.next;
        }
        return null;
    }
    private int length(ListNode node) {
        int len=0;
        while(node!=null){
            node=node.next;
            len++;
        }
        return len;
    }
}

第二个是用值的差 O(2(m+n))

public class Solution {
    public ListNode getIntersectionNode(ListNode headA, ListNode headB) {
        if(headA==null||headB==null) return null;
        ListNode a=headA;
        ListNode b=headB;
        int sumA=0;
        int sumB=0;
        int ta=0;
        int tb=0;
        while(a!=null||b!=null){
            if(a!=null){
                sumA+=a.val;
                a=a.next;
                ta++;
            }
            if(b!=null){
                sumB+=b.val;
                b=b.next;
                tb++;
            }
            
        }
        a=headA;
        b=headB;
        while(sumA>0&&sumB>0){
            if(sumA==sumB&&ta==tb) return a;
            else if(sumA>sumB){
                sumA-=a.val;
                a=a.next;
                ta--;
            }
            else{
                sumB-=b.val;
                b=b.next;
                tb--;
            }
        }
        return null;
    }
}