两个链表的公共值

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现有两个升序链表，且链表中均无重复元素。请设计一个高效的算法，打印两个链表的公共值部分。

给定两个链表的头指针headA和headB，请返回一个vector，元素为两个链表的公共部分。请保证返回数组的升序。两个链表的元素个数均小于等于500。保证一定有公共值

测试样例：

{1,2,3,4,5,6,7},{2,4,6,8,10}

返回：[2.4.6]

代码如下：

package lianbiao;import java.util.ArrayList;import java.util.List;public class lianbiaogonggongzhi {public static void main(String[] args) {ListNode headA = new ListNode(3);ListNode headA2 = new ListNode(4);ListNode headA3 = new ListNode(5);headA.next = headA2;headA2.next = headA3;ListNode headB = new ListNode(4);ListNode headB2 = new ListNode(5);ListNode headB3 = new ListNode(6);ListNode headB4 = new ListNode(7);headB.next = headB2;headB2.next = headB3;headB3.next = headB4;int[] arr = lianbiaogonggongzhi.findCommonParts(headA, headB);for (int i = 0; i < arr.length; i++) {              System.out.print(arr[i]+" ");}}public static int[] findCommonParts(ListNode headA, ListNode headB) {// write code hereList<Integer> list = new ArrayList<Integer>();while (headA != null && headB != null) {if (headA.val < headB.val) {headA = headA.next;} else if (headA.val > headB.val) {headB = headB.next;} else {list.add(headA.val);headA = headA.next;headB = headB.next;}}int[] arr = new int[list.size()];for (int i = 0; i < arr.length; i++) {arr[i] = list.get(i);}return arr;}static class ListNode {int val = 0;ListNode next = null;ListNode(int val) {this.val = val;}}}

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