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LeetCode-合并两个有序链表
Table of Contents
- 1. Easy-合并两个有序链表
- 1.1. 题目描述
- 2. 自己的答案
- 2.1. 思路
- 2.2. 代码
- 2.3. 反思
- 2.4. 优化
1 Easy-合并两个有序链表
1.1 题目描述
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
2 自己的答案
2.1 思路
- 如果l1或者l2其中一个为null,则返回null.
- 用res变量作为返回结果,并创建一个别名为tmp的变量
- 当l1当前值大于l2当前值,tmp的值为l2当前的值,并且l2移动到下一个节点.l1小于l2的情况类似.
- tmp的下一个变量初始化为0,并且tmp移动到下一个节点
- 直到l1或者l2其中一个到节点的尾部,循环结束,将未到达尾部的链表剩余部分归并到tmp.next,返回res
2.2 代码
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
ListNode res = new ListNode(0);
ListNode tmp = res;
// 有一个方为空,则返回另一方
if (l1 == null)
return l2;
if (l2 == null)
return l1;
// 开始遍历l1和l2
while (true) {
// l1链表的值大于l2链表的值
if (l1.val > l2.val) {
tmp.val = l2.val;
// l2为下一个节点
l2 = l2.next;
} else {
tmp.val = l1.val;
// l1为下一个节点
l1 = l1.next;
}
// 如果l1或者l2其中一个到了节点尾部,将另一个链表接到tmp上并直接返回
if (l1 == null) {
// l1没有元素,把l2的元素加到tmp结尾并返回
tmp.next = l2;
return res;
}
if (l2 == null) {
// l2没有元素,把l1的元素加到tmp结尾并返回
tmp.next = l1;
return res;
}
// 初始化新的节点,赋初值为0
tmp.next = new ListNode(0);
tmp = tmp.next;
}
}
}
2.3 反思
- 为tmp初始化为0的步骤是冗余的步骤,增加了空间复杂度
- 对Java别名使用的不熟悉,函数传参,赋值等操作的均是按值传递,传递的是变量的地址
- 尽可能不用
while(true),容易出现死循环
2.4 优化
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
class Solution {
public ListNode mergeTwoLists(ListNode l1, ListNode l2) {
// 初始化一个res,值为0
ListNode res = new ListNode(0);
// 用tmp别名替代res,以免改变返回结果的指向
ListNode tmp = res;
// 有一个方为空,则返回另一方
if (l1 == null)
return l2;
if (l2 == null)
return l1;
// 开始遍历l1和l2
while (l1 != null && l2 != null) {
// l1链表的值大于l2链表的值
if (l1.val > l2.val) {
// tmp是别名,指向当前l2,l2改变的时候当前tmp值不会改变
tmp.next = l2;
// l2为下一个节点
l2 = l2.next;
} else {
// tmp是别名,指向当前l1,l1改变的时候当前tmp的值不会改变
tmp.next = l1;
// l1为下一个节点
l1 = l1.next;
}
tmp = tmp.next;
}
// 如果l1或者l2其中一个到了节点尾部,将另一个链表接到tmp上并直接返回
tmp.next = (l1 == null ? l2 : l1);
return res.next;
}
}
Date: 2018-10-29 23:23
Author: devinkin
Created: 2018-10-29 一 23:43
原文地址：https://www.cnblogs.com/devinkin/p/9874116.html