[剑指offer] 合并两个排序的链表

题目描述

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。

解题思路

两种解法：递归和非递归

参考代码

/*
public class ListNode {
    int val;
    ListNode next = null;

    ListNode(int val) {
        this.val = val;
    }
}*/
//递归解法
public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        if(list1 == null)
            return list2;
        else if(list2 == null)
            return list1;
        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);
        }
        return mergehead;
    }
}
//非递归解法
public class Solution {
    public ListNode Merge(ListNode list1,ListNode list2) {
        if(list1 == null)
            return list2;
        else if(list2 == null)
            return list1;
        ListNode mergehead = null;
        if(list1.val <= list2.val){
            mergehead = list1;
            list1 = list1.next;
        }else{
            mergehead = list2;
            list2 = list2.next;
        }
        ListNode cur = mergehead;
        while(list1 != null && list2 != null){
            if(list1.val <= list2.val){
                cur.next = list1;
                list1 = list1.next;
            }else{
                cur.next = list2;
                list2 = list2.next;
            }
            cur = cur.next;
        }
        if(list1 == null)
            cur.next = list2;
        else if(list2 == null)
            cur.next = list1;
        return mergehead;
    }
}

        

          
        
        
        

          

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