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 pHead2;
        if (pHead2 == null) return pHead1;
        //递归最简单: 相当于每次取一个最小的节点
        if(pHead1.val<=pHead2.val){
            pHead1.next = Merge(pHead1.next, pHead2);
            return pHead1;
        }else{
            pHead2.next = Merge(pHead2.next, pHead1);
            return pHead2;
        }
    }

    public void printNode(ListNode node) {
        StringBuilder sb = new StringBuilder();
        sb.append("node: ");
        while (node != null) {
            sb.append(node.val);
            sb.append("->");
            node = node.next;
        }
        System.out.println(sb.toString());
    }

    public ListNode find(ListNode a, int v) {
        //a是递增链表且a.val<=v，寻找>V前一个节点
        ListNode te = a;
        //直接拿next比对
        while (a.next != null) {
            if (a.next.val <= v) {
                a = a.next;
            } else {
                return a ;
            }
        }
        //遍历到最后都>v, 返回最后一个节点a即可
        return a;
    }
}