[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

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/*
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;
}
}