【面试题17】合并两个排序的链表

来源：互联网发布：fm2017梅西数据编辑：程序博客网时间：2024/06/05 11:16

输入两个单调递增的链表，输出两个链表合成后的链表，当然我们需要合成后的链表满足单调不减规则。

递归实现：

/*
public class ListNode {
    int val;
    ListNode next = null;
 
    ListNode(int val) {
        this.val = val;
    }
}*/
public class Solution {
    publicListNode Merge(ListNode list1,ListNode list2) {
        if(list1 ==null)
            returnlist2;
        elseif(list2 == null)
            returnlist1;
         
        ListNode mergehead =null;
        if(list1.val < list2.val) {
            mergehead = list1;
            mergehead.next = Merge(list1.next, list2);
        }else {
            mergehead = list2;
            mergehead.next = Merge(list1, list2.next);
        }
        returnmergehead;
    }
}

非递归方式：

public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        if (list1 ==null)
            return list2;
        if (list2 ==null)
            return list1;
        ListNode head = null;
        ListNode tmp= null;
        if (list1.val < list2.val)
            {
            head = list1;
                list1 = list1.next;
        }
        else
        {
            head = list2;
                list2 = list2.next;
        }
        tmp = head;
        while (list1 != null && list2 != null)
            {
            if (list1.val < list2.val)
            {

            tmp.next = list1;
                list1 = list1.next;
                tmp = tmp.next;

        }
        else
        {
           tmp.next = list2;
                list2 = list2.next;
            tmp = tmp.next;
        }
        }

         while (list1 != null || list2 != null)
            {
        if (list1 ==null && list2 != null)
            {
            tmp.next = list2;
                list2 = list2.next;
            tmp = tmp.next;
        }

         if (list2 ==null && list1 != null)
            {
            tmp.next = list1;
                list1 = list1.next;
            tmp = tmp.next;
        }
         }
        return head;

    }
}

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