剑指offer 17. 合并两个排序的链表

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// 题目：合并两个增序链表// 解法1：非递归的方法public class Main {public static void main(String[] args) {ListNode n1 = new ListNode();ListNode n2 = new ListNode();ListNode n3 = new ListNode();ListNode n4 = new ListNode();ListNode t1 = new ListNode();ListNode t2 = new ListNode();ListNode t3 = new ListNode();ListNode t4 = new ListNode();n1.data = 1;n2.data = 3;n3.data = 5;n4.data = 7;t1.data = 2;t2.data = 4;t3.data = 6;t4.data = 8;n1.next = n2;n2.next = n3;n3.next = n4;t1.next = t2;t2.next = t3;t3.next = t4;ListNode result = mergeList(n1,t1);while(result!=null){System.out.println(result.data);result = result.next;}}public static ListNode mergeList(ListNode node1, ListNode node2) {if(node1 == null){return node2;}if(node2 == null){return node1;}ListNode result;ListNode temp;//返回链表中每次操作的结点if(node1.data<node2.data){//确定第一个返回的结点result = node1;temp = node1;node1 = node1.next;}else{result = node2;temp = node2;node2 = node2.next;}while(node1!=null && node2!=null){//一直比较，知道一个链表被遍历完if(node1.data<node2.data){temp.next = node1;temp = temp.next;node1 = node1.next;}else{temp.next = node2;temp = temp.next;node2 = node2.next;}}while(node1!=null){//把没有遍历完的部分连接到结果之后temp.next = node1;break;}while(node2!=null){temp.next = node2;break;}return result;}}// 解法2：递归的方法public class Main {public static void main(String[] args) {ListNode n1 = new ListNode();ListNode n2 = new ListNode();ListNode n3 = new ListNode();ListNode n4 = new ListNode();ListNode t1 = new ListNode();ListNode t2 = new ListNode();ListNode t3 = new ListNode();ListNode t4 = new ListNode();n1.data = 1;n2.data = 3;n3.data = 5;n4.data = 7;t1.data = 2;t2.data = 4;t3.data = 6;t4.data = 8;n1.next = n2;n2.next = n3;n3.next = n4;t1.next = t2;t2.next = t3;t3.next = t4;ListNode result = mergeList(n1,t1);while(result!=null){System.out.println(result.data);result = result.next;}}public static ListNode mergeList(ListNode node1, ListNode node2) {if(node1 == null){//如果一个遍历完了就返回另一个return node2;}if(node2 == null){return node1;}ListNode result;if(node1.data<node2.data){//找到当前节点比较小的作为返回result = node1;node1 = node1.next;}else{result = node2;node2 = node2.next;}result.next = mergeList(node1,node2);//递归的判断返回结点的下一节点return result;}}
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