160. Intersection of Two Linked Lists（js）

160. Intersection of Two Linked Lists

Write a program to find the node at which the intersection of two singly linked lists begins.

For example, the following two linked lists:

begin to intersect at node c1.

Example 1:

Input: intersectVal = 8, listA = [4,1,8,4,5], listB = [5,0,1,8,4,5], skipA = 2, skipB = 3
Output: Reference of the node with value = 8
Input Explanation: The intersected node‘s value is 8 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [4,1,8,4,5]. From the head of B, it reads as [5,0,1,8,4,5]. There are 2 nodes before the intersected node in A; There are 3 nodes before the intersected node in B.

Example 2:

Input: intersectVal = 2, listA = [0,9,1,2,4], listB = [3,2,4], skipA = 3, skipB = 1
Output: Reference of the node with value = 2
Input Explanation: The intersected node‘s value is 2 (note that this must not be 0 if the two lists intersect). From the head of A, it reads as [0,9,1,2,4]. From the head of B, it reads as [3,2,4]. There are 3 nodes before the intersected node in A; There are 1 node before the intersected node in B.

Example 3:

Input: intersectVal = 0, listA = [2,6,4], listB = [1,5], skipA = 3, skipB = 2
Output: null
Input Explanation: From the head of A, it reads as [2,6,4]. From the head of B, it reads as [1,5]. Since the two lists do not intersect, intersectVal must be 0, while skipA and skipB can be arbitrary values.
Explanation: The two lists do not intersect, so return null.

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.

题意：给出两个链表，求出相交节点的值，若不相交则返回null

代码如下：

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */

/**
 * @param {ListNode} headA
 * @param {ListNode} headB
 * @return {ListNode}
 */
// 基本思路：
// 1.比较两个链表的长度
// 2.移动指针节点，较长的先移动，使得两链表长度相等时再一起移动
// 3.当两个指针节点指向同一节点时即交点
var getIntersectionNode = function(headA, headB) {
        if(!headA || !headB) return null;
        let alen=getLength(headA);
        let blen=getLength(headB);
        if(alen>blen){
            for(let i=0;i<alen-blen;i++){   headA=headA.next;}
        }else{
            for(let i=0;i<blen-alen;i++){   headB=headB.next;}
        }

        while(headA && headB ){
            if(headA==headB){
                return headA;
            }
            headA=headA.next;
            headB=headB.next;

        }
        return null;
};
//获取链表长度
var getLength = function(node){
    let len=0;
    while(node){
        node=node.next;
        len++;
    }
    return len;
}

原文地址：https://www.cnblogs.com/xingguozhiming/p/11336371.html