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feat: 0002.Add-Two-Numbers & 0143.Reorder-List
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‎Leetcode/0143.Reorder-List/README.md

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---
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title: 143. Reorder List
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subtitle: "https://leetcode.com/problems/reorder-list/description/"
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date: 2024-03-03T16:53:00+08:00
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lastmod: 2024-03-03T16:53:00+08:00
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draft: false
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author: "Kimi.Tsai"
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authorLink: "https://kimi0230.github.io/"
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description: "0143.Reorder-List"
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license: ""
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images: []
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tags: [LeetCode, Go, /Medium, 143. Reorder List, Amazon,Microsoft,Adobe,Facebook,Bloomberg,Linked List,Two Pointers,Stack, Recursion]
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categories: [LeetCode]
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featuredImage: ""
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featuredImagePreview: ""
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hiddenFromHomePage: false
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hiddenFromSearch: false
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twemoji: false
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lightgallery: true
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ruby: true
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fraction: true
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fontawesome: true
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linkToMarkdown: false
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rssFullText: false
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toc:
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enable: true
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auto: true
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code:
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copy: true
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maxShownLines: 200
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math:
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enable: false
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# ...
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mapbox:
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# ...
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share:
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enable: true
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# ...
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comment:
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enable: true
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# ...
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library:
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css:
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# someCSS = "some.css"
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# located in "assets/"
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# Or
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# someCSS = "https://cdn.example.com/some.css"
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js:
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# someJS = "some.js"
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# located in "assets/"
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# Or
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# someJS = "https://cdn.example.com/some.js"
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seo:
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images: []
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# ...
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---
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# [143. Reorder List](https://leetcode.com/problems/reorder-list/description/)
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## 題目
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## 題目大意
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## 解題思路
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## Big O
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* 時間複雜 : ``
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* 空間複雜 : ``
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## 來源
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* https://leetcode.com/problems/reorder-list/description/
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* https://leetcode.cn/problems/
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## 解答
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https://github.com/kimi0230/LeetcodeGolang/blob/master/Leetcode/0143.Reorder-List/main.go
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```go
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package reorderlist
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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// 時間複雜 O(), 空間複雜 O()
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// 先用快慢指針找出Linked list的中間點
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// 然後把Linked list分成兩半
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// 再把後半的Linked list反轉
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// 再把兩半的Linked list merge 起來
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func reorderList(head *ListNode) {
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mid := middleNode(head)
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// 2.反轉中間節點的下一個節點
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right := resverseNode(mid.Next)
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mid.Next = nil
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left := head
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mergeNode(left, right)
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}
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// [876. Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/)
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func middleNode(head *ListNode) *ListNode {
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slow, fast := head, head
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for fast != nil && fast.Next != nil {
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slow = slow.Next
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fast = fast.Next.Next
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}
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return slow
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}
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// [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)
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func resverseNode(head *ListNode) *ListNode {
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var pre *ListNode
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for head != nil {
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tmp := head.Next
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head.Next = pre
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pre = head
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head = tmp
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}
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return pre
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}
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func mergeNode(l1, l2 *ListNode) {
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lcur, rcur := l1, l2
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for lcur != nil && rcur != nil {
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lcur.Next, rcur.Next, lcur, rcur = rcur, lcur.Next, lcur.Next, rcur.Next
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}
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}
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```
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## Benchmark
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```sh
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```

‎Leetcode/0143.Reorder-List/main.go

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package reorderlist
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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// 時間複雜 O(), 空間複雜 O()
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// 先用快慢指針找出Linked list的中間點
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// 然後把Linked list分成兩半
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// 再把後半的Linked list反轉
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// 再把兩半的Linked list merge 起來
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func reorderList(head *ListNode) {
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mid := middleNode(head)
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// 2.反轉中間節點的下一個節點
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right := resverseNode(mid.Next)
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mid.Next = nil
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left := head
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mergeNode(left, right)
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}
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// [876. Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/)
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func middleNode(head *ListNode) *ListNode {
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slow, fast := head, head
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for fast != nil && fast.Next != nil {
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slow = slow.Next
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fast = fast.Next.Next
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}
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return slow
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}
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// [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)
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func resverseNode(head *ListNode) *ListNode {
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var pre *ListNode
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for head != nil {
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tmp := head.Next
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head.Next = pre
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pre = head
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head = tmp
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}
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return pre
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}
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func mergeNode(l1, l2 *ListNode) {
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lcur, rcur := l1, l2
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for lcur != nil && rcur != nil {
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lcur.Next, rcur.Next, lcur, rcur = rcur, lcur.Next, lcur.Next, rcur.Next
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}
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}

‎Leetcode/0143.Reorder-List/main_test.go

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package
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import "testing"
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var tests = []struct {
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arg1 string
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want int
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}{
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{
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"bbbab",
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4,
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},
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}
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func TestLongestPalindromeSubseq(t *testing.T) {
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for _, tt := range tests {
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// if got := ReverseList(tt.arg1); !reflect.DeepEqual(got, tt.want) {
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if got := LongestPalindromeSubseq(tt.arg1); got != tt.want {
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t.Errorf("got = %v, want = %v", got, tt.want)
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}
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}
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}
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func BenchmarkLongestPalindromeSubseq(b *testing.B) {
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b.ResetTimer()
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for i := 0; i < b.N; i++ {
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LongestPalindromeSubseq(tests[0].arg1)
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}
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}
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/*
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go test -benchmem -run=none LeetcodeGolang/Leetcode/0354.Russian-Doll-Envelopes -bench=.
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*/

‎Leetcode_labuladong/143.重排链表.go

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// 時間複雜度:
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// 空間複雜度:
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/*
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* @lc app=leetcode.cn id=143 lang=golang
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*
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* [143] 重排链表
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*/
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// @lc code=start
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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// 先用快慢指針找出Linked list的中間點
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// 然後把Linked list分成兩半
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// 再把後半的Linked list反轉
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// 再把兩半的Linked list merge 起來
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func reorderList(head *ListNode) {
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mid := middleNode(head)
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// 2.反轉中間節點的下一個節點
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right := resverseNode(mid.Next)
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mid.Next = nil
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left := head
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mergeNode(left, right)
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}
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// [876. Middle of the Linked List](https://leetcode.com/problems/middle-of-the-linked-list/)
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func middleNode(head *ListNode) *ListNode {
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slow, fast := head, head
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for fast != nil && fast.Next != nil {
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slow = slow.Next
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fast = fast.Next.Next
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}
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return slow
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}
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// [206. Reverse Linked List](https://leetcode.com/problems/reverse-linked-list/)
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func resverseNode(head *ListNode) *ListNode {
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var pre *ListNode
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for head != nil {
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tmp := head.Next
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head.Next = pre
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pre = head
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head = tmp
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}
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return pre
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}
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func mergeNode(l1, l2 *ListNode) {
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lcur, rcur := l1, l2
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for lcur != nil && rcur != nil {
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lcur.Next, rcur.Next, lcur, rcur = rcur, lcur.Next, lcur.Next, rcur.Next
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}
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}
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// @lc code=end
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‎Leetcode_labuladong/2.两数相加.go

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// 時間複雜度:
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// 空間複雜度:
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/*
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* @lc app=leetcode.cn id=2 lang=golang
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*
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* [2] 两数相加
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*/
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// @lc code=start
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/**
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* Definition for singly-linked list.
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* type ListNode struct {
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* Val int
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* Next *ListNode
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* }
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*/
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// 遍歷 l1跟 l2. 講兩個list的val相加, 並且記錄進位的值給next使用
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// 最後如果 carry 還有的話, 需要產生一個新的節點
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func addTwoNumbers(l1 *ListNode, l2 *ListNode) *ListNode {
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var result, tail *ListNode
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carry := 0
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for l1 != nil || l2 != nil {
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n1, n2 := 0, 0
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if l1 != nil {
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n1 = l1.Val
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l1 = l1.Next
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}
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if l2 != nil {
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n2 = l2.Val
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l2 = l2.Next
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}
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sum := n1 + n2 + carry
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sum, carry = sum%10, sum/10
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if result == nil {
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result = &ListNode{Val: sum}
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tail = result
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} else {
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tail.Next = &ListNode{Val: sum}
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tail = tail.Next
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}
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}
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if carry > 0 {
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tail.Next = &ListNode{Val: carry}
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}
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return result
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}
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// @lc code=end
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