剑指offer第二版-25.合并两个排序的链表

本系列导航：剑指offer（第二版）java实现导航帖

面试题25：合并两个排序的链表

题目要求：
输入两个递增排序的链表，要求合并后保持递增。

解题思路：
这个题目是二路链表归并排序的一部分，或者说是最关键的归并函数部分。熟悉归并排序的话做这个题应该很容易。思路很简单，注意链表的next指针，初始条件，结束条件即可。

package structure;
/**
 * Created by ryder on 2017/6/13.
 */
public class ListNode {
    public T val;
    public ListNode next;
    public ListNode(T val){
        this.val = val;
        this.next = null;
    }
    @Override
    public String toString() {
        StringBuilder ret = new StringBuilder();
        ret.append("[");
        for(ListNode cur = this;;cur=cur.next){
            if(cur==null){
                ret.deleteCharAt(ret.lastIndexOf(" "));
                ret.deleteCharAt(ret.lastIndexOf(","));
                break;
            }
            ret.append(cur.val);
            ret.append(", ");
        }
        ret.append("]");
        return ret.toString();
    }
}

package chapter3;
import structure.ListNode;
/**
 * Created by ryder on 2017/7/14.
 * 合并两个排序的链表
 */
public class P145_MergeSortedLists {
    public static ListNode merge(ListNode head1,ListNode head2){
        if(head1==null)
            return head2;
        if(head2==null)
            return head1;
       ListNode index1 = head1;
       ListNode index2 = head2;
       ListNode index = null;
       if(index1.val head1 = new ListNode<>(1);
        head1.next= new ListNode<>(3);
        head1.next.next = new ListNode<>(5);
        head1.next.next.next = new ListNode<>(7);
        ListNode head2 = new ListNode<>(2);
        head2.next= new ListNode<>(4);
        head2.next.next = new ListNode<>(6);
        head2.next.next.next = new ListNode<>(8);
        System.out.println(head1);
        System.out.println(head2);
        ListNode head =merge(head1,head2);
        System.out.println(head);
    }
}

运行结果

[1, 3, 5, 7]
[2, 4, 6, 8]
[1, 2, 3, 4, 5, 6, 7, 8]