题解 | #合并两个排序的链表#

import java.util.*;

/*
 * public class ListNode {
 *   int val;
 *   ListNode next = null;
 *   public ListNode(int val) {
 *     this.val = val;
 *   }
 * }
 */

public class Solution {
    /**
     * 代码中的类名、方法名、参数名已经指定，请勿修改，直接返回方法规定的值即可
     *
     *
     * @param pHead1 ListNode类
     * @param pHead2 ListNode类
     * @return ListNode类
     */
    public ListNode Merge (ListNode pHead1, ListNode pHead2) {
        // write code here
        if (pHead1 == null) {
            return pHead1;
        }
        ListNode cur = new ListNode(0);
        ListNode head = cur;
        while (pHead1 != null && pHead2 != null) {
            if (pHead1.val < pHead2.val) {
                cur.val = pHead1.val;
                pHead1 = pHead1.next;
            } else {
                cur.val = pHead2.val;
                pHead2 = pHead2.next;
            }
            cur.next = new ListNode(0);
            cur = cur.next;
        }
        while (pHead1 != null) {
            cur.val = pHead1.val;
            pHead1 = pHead1.next;
            if (pHead1 != null) {
                cur.next = new ListNode(0);
                cur = cur.next;
            } else {
                break;
            }
        }

        while (pHead2 != null) {
            cur.val = pHead2.val;
            pHead2 = pHead2.next;
            if (pHead2 != null) {
                cur.next = new ListNode(0);
                cur = cur.next;
            } else {
                break;
            }
        }

        return head;
    }
}

双指针法，考虑到两个链表长度是一致的，一个为0则另一个也为0,边界处理返回哪个链表都行。后面主要考虑三种情况，两个链表都遍历完成、pHead1遍历完pHead2没遍历完、pHead2遍历完pHead1没遍历完。后续用比较简单的思路处理一下。

其实关于这些有序数据结构合并，基本就是三种思路：先合并再排序、双指针、逆双指针。本人水平有限，就先用双指针解题。