剑指offer第二版-25.合并两个排序的链表

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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<T> {
    public T val;
    public ListNode<T> 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<Integer> merge(ListNode<Integer> head1,ListNode<Integer> head2){
        if(head1==null)
            return head2;
        if(head2==null)
            return head1;
       ListNode<Integer> index1 = head1;
       ListNode<Integer> index2 = head2;
       ListNode<Integer> index = null;
       if(index1.val<index2.val) {
           index = index1;
           index1 = index1.next;
       }
       else {
           index = index2;
           index2 = index2.next;
       }
       while(index1!=null && index2!=null){
           if(index1.val<index2.val) {
               index.next = index1;
               index = index.next;
               index1 = index1.next;
           }
           else {
               index.next = index2;
               index = index.next;
               index2 = index2.next;
           }
       }
       if(index1!=null)
           index.next = index1;
       else
           index.next = index2;
       return head1.val<head2.val?head1:head2;
    }
    public static void main(String[] args){
        ListNode<Integer> head1 = new ListNode<>(1);
        head1.next= new ListNode<>(3);
        head1.next.next = new ListNode<>(5);
        head1.next.next.next = new ListNode<>(7);
        ListNode<Integer> 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<Integer> 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]

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