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添加 problem 92
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Algorithms/92.Reverse Linked List II/README.md
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23
Algorithms/92.Reverse Linked List II/README.md
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# [92. Reverse Linked List II](https://leetcode.com/problems/reverse-linked-list-ii/)
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## 题目
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Reverse a linked list from position m to n. Do it in one-pass.
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Note: 1 ≤ m ≤ n ≤ length of list.
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Example:
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```
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Input: 1->2->3->4->5->NULL, m = 2, n = 4
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Output: 1->4->3->2->5->NULL
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```
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## 题目大意
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给定 2 个链表中结点的位置 m, n,反转这个两个位置区间内的所有结点。
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由于有可能整个链表都被反转,所以构造一个新的头结点指向当前的头。之后的处理方法是:找到第一个需要反转的结点的前一个结点 p,从这个结点开始,依次把后面的结点用“头插”法,插入到 p 结点的后面。循环次数用 n-m 来控制。
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这一题结点可以原地变化,更改各个结点的 next 指针就可以。不需要游标 p 指针。因为每次逆序以后,原有结点的相对位置就发生了变化,相当于游标指针已经移动了,所以不需要再有游标 p = p.Next 的操作了。
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package leetcode
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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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func reverseBetween(head *ListNode, m int, n int) *ListNode {
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if head == nil || m >= n {
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return head
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}
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newHead := &ListNode{Val: 0, Next: head}
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pre := newHead
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for count := 0; pre.Next != nil && count < m-1; count++ {
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pre = pre.Next
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}
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if pre.Next == nil {
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return head
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}
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cur := pre.Next
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for i := 0; i < n-m; i++ {
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tmp := pre.Next
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pre.Next = cur.Next
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cur.Next = cur.Next.Next
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pre.Next.Next = tmp
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}
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return newHead.Next
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}
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package leetcode
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import (
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"fmt"
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"testing"
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)
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type question92 struct {
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para92
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ans92
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}
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// para 是参数
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// one 代表第一个参数
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type para92 struct {
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one []int
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m, n int
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}
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// ans 是答案
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// one 代表第一个答案
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type ans92 struct {
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one []int
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}
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func Test_Problem92(t *testing.T) {
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qs := []question92{
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question92{
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para92{[]int{1, 2, 3, 4, 5}, 2, 4},
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ans92{[]int{1, 4, 3, 2, 5}},
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},
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question92{
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para92{[]int{1, 2, 3, 4, 5}, 2, 2},
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ans92{[]int{1, 2, 3, 4, 5}},
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},
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question92{
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para92{[]int{1, 2, 3, 4, 5}, 1, 5},
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ans92{[]int{5, 4, 3, 2, 1}},
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},
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question92{
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para92{[]int{1, 2, 3, 4, 5, 6}, 3, 4},
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ans92{[]int{1, 2, 4, 3, 5, 6}},
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},
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question92{
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para92{[]int{3, 5}, 1, 2},
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ans92{[]int{5, 3}},
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},
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question92{
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para92{[]int{3}, 3, 5},
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ans92{[]int{3}},
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},
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}
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fmt.Printf("------------------------Leetcode Problem 92------------------------\n")
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for _, q := range qs {
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_, p := q.ans92, q.para92
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fmt.Printf("【input】:%v 【output】:%v\n", p, L2s(reverseBetween(S2l(p.one), p.m, p.n)))
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}
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fmt.Printf("\n\n\n")
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}
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