有序链表的构造
class ListNode{
int val;
ListNode nextNode;
// 构造函数
ListNode(int val){
this.val=val;
this.nextNode=null;
}
}
public static ListNode buildListNode(int [] list){
//创建3个临时的ListNode
ListNode first=null,last=null,newNode;
for(int i=0;i<list.length;i++){
newNode=new ListNode(list[i]);
if(first==null){
first=newNode;
last=newNode;
}else{
last.nextNode=newNode;
last=newNode;
}
}
return first;
}
有序链表的简单打印:
int[] a=new int[]{1,5,6};
ListNode alist=buildListNode( a);
ListNode testnode = alist;
while(testnode != null) {
System.out.println("-->" + testnode.val);
testnode=testnode.nextNode;
}
-->1-->5-->6
面试题1:
将两个有序链表合并为一个新的有序链表并返回。新链表是通过拼接给定的两个链表的所有节点组成的。
示例:
输入:1->2->4, 1->3->4
输出:1->1->2->3->4->4
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1 == null) return l2;
if (l2 == null) return l1;
ListNode head = null;
if (l1.val <= l2.val){
head = l1;
head.next = mergeTwoLists(l1.next, l2);
} else {
head = l2;
head.next = mergeTwoLists(l1, l2.next);
}
return head;
}
}