合并两个有序链表(Java实现)
题目: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
解法一:常规解法
package Day47;
/**
* @Author Zhongger
* @Description 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。
* @Date 2020.3.20
*/
public class MergeTwoListsSolution {
public static void main(String[] args) {
ListNode l1_1 = new ListNode(1);
ListNode l1_2 = new ListNode(2);
ListNode l1_3 = new ListNode(4);
l1_1.next=l1_2;
l1_2.next=l1_3;
ListNode l2_1 = new ListNode(1);
ListNode l2_2 = new ListNode(3);
ListNode l2_3 = new ListNode(4);
l2_1.next=l2_2;
l2_2.next=l2_3;
MergeTwoListsSolution solution = new MergeTwoListsSolution();
ListNode mergeTwoLists = solution.mergeTwoLists(l1_1, l2_1);
ListNode temp=mergeTwoLists;
while (temp!=null){
System.out.println(temp);
temp=temp.next;
}
}
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
if (l1==null&&l2==null){
return null;
}
ListNode resList=new ListNode(0);//头结点
ListNode temp=resList;//辅助节点
while (l1!=null&&l2!=null){
if (l1.val<l2.val){
temp.next=new ListNode(l1.val);
l1=l1.next;
}else {
temp.next=new ListNode(l2.val);
l2=l2.next;
}
temp=temp.next;
}
if (l1==null){
temp.next=l2;
}
if (l2==null){
temp.next=l1;
}
return resList.next;
}
}
class ListNode {
int val;
ListNode next;
public ListNode(int val) {
this.val = val;
}
@Override
public String toString() {
return val+" ";
}
}
解法二:递归解法
/**
* 递归求解
* @param l1
* @param l2
* @return
*/
public ListNode mergeTwoLists2(ListNode l1, ListNode l2) {
if (l1 == null) {
return l2;
}
else if (l2 == null) {
return l1;
}
else if (l1.val < l2.val) {
l1.next = mergeTwoLists(l1.next, l2);
return l1;
}
else {
l2.next = mergeTwoLists(l1, l2.next);
return l2;
}
}
关于递归求解的方法,可以查看该链接。https://leetcode-cn.com/problems/merge-two-sorted-lists/solution/yi-kan-jiu-hui-yi-xie-jiu-fei-xiang-jie-di-gui-by-/