剑指offer --合并链表

题目: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。

解法://递归解法

public class MixLink {

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

ListNode(int val) {
this.val = val;
}
}*/
public class Solution {
public ListNode Merge(ListNode list1,ListNode list2) {
if(list1==null){
return list2;
}
if(list2==null){
return list1;
}
ListNode list3 = null;

if(list1.val<list2.val){
list3 = list1;
list3.next = Merge(list1.next,list2);
}else if(list1.val>=list2.val){
list3 = list2;
list3.next = Merge(list1,list2.next);
}
return list3;
}
}

解法: 非递归

       while(list1!=null&&list2!=null){
              
             if(list1.val<list2.val){
                 if(list3==null){
                      list3 = list1;
                 }else{
                     list3.next = list1;
                     list3 = list3.next; 
                 }
                 list1 = list1.next;
             }else if(list1.val>list2.val){
                 if(list3 == null){
                      list3 = list2;
                 }else{
                     list3.next = list2;
                     list3 = list3.next;
                 }
                 list2 = list2.next;
             }
              
         }
         if(list1==null){
             list3.next= list2;
         }
         if(list2 == null){
             list3.next = list1;
         }
         return list3;
原文地址:https://www.cnblogs.com/tandy/p/5443872.html