版权声明:禁止侵权,打击盗版! https://blog.csdn.net/ChenGX1996/article/details/82120911
//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
/*
public class ListNode {
int val;
ListNode next = null;
ListNode(int val) {
this.val = val;
}
}*/
public class Solution {
public ListNode Merge(ListNode list1,ListNode list2) {
//假如某一或两个链表为空
if(list1 == null){
return list2;
}
if(list2 == null){
return list1;
}
if(list1 == null && list2 == null){
return null;
}
ListNode pNode = null;
if(list1.val < list2.val){
pNode = list1;
//运用递归
pNode.next = Merge(list1.next,list2);
}else{
pNode = list2;
pNode.next = Merge(list1,list2.next);
}
return pNode;
}
}