一、题目1
二、思路
三、代码
public class T0021 {
public static void main(String[] args) {
int[] nums1 = { 1, 3, 5 };
int[] nums2 = { 2, 4, 6 };
ListNode l1 = new ListNode(nums1[0]);
ListNode node = l1;
for ( int i = 1; i < nums1.length; i++ ){
node.next = new ListNode(nums1[i]);
node = node.next;
}
ListNode l2 = new ListNode(nums2[0]);
node = l2;
for ( int i = 1; i < nums2.length; i++ ){
node.next = new ListNode(nums2[i]);
node = node.next;
}
ListNode head = mergeTwoLists( l1, l2 );
while ( head != null ){
System.out.println( head.val );
head = head.next;
}
}
public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode head = new ListNode(0); //头结点
ListNode node = head; //用于迭代的节点
//只要有剩余的节点就继续进行循环
while ( l1 != null || l2 != null ){
//如果两个节点均不为空,将较小的节点进行保存,较大的不管
if ( l1 != null && l2 != null ){
if ( l1.val < l2.val ){
node.next = l1;
l1 = l1.next;
}else {
node.next = l2;
l2 = l2.next;
}
//仅l1不为空,将l1保存后退出
}else if ( l1 != null && l2 == null ){
node.next = l1;
break;
//仅l2不为空,将l2保存后退出
}else{
node.next = l2;
break;
}
node = node.next;
}
//头结点的下一个节点才是有效值
return head.next;
}
}
//Definition for singly-linked list.
class ListNode {
int val;
ListNode next;
ListNode(int x) { val = x; }
}
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/merge-two-sorted-lists
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