【CT】LeetCode手撕—92. 反转链表 II

题目

原题连接：92. 反转链表 II

1- 思路

模式识别：翻转给定 left 和 right 固定区间的链表 ——> ①虚拟头结点 + ②三个指针方式实现

2- 实现

⭐92. 反转链表 II——题解思路

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class Solution {public ListNode reverseBetween(ListNode head, int left, int right) {ListNode dummyHead = new ListNode(-1);dummyHead.next = head;ListNode pre = dummyHead;for(int i = 1 ; i < left;i++){pre = pre.next;}ListNode cur = pre.next;ListNode next = null;for(int i = 0 ; i < right - left;i++){next = cur.next;cur.next = next.next;next.next = pre.next;pre.next = next;}return dummyHead.next;}
}

3- ACM实现

public class reverse {static class ListNode{int val;ListNode next;ListNode(){}ListNode(int x){val = x;}}public static ListNode reverseList(ListNode head,int left,int right){ListNode dummyHead = new ListNode(-1);dummyHead.next = head;ListNode pre = dummyHead;for(int i = 1 ; i < left;i++){pre = pre.next;}ListNode cur = pre.next;ListNode next =null;for(int i = 0 ; i < right-left;i++){next = cur.next;cur.next = next.next;next.next = pre.next;pre.next = next;}return dummyHead.next;}public static void main(String[] args) {Scanner sc = new Scanner(System.in);System.out.println("输入链表长度");int n =sc.nextInt();ListNode head = null,tail = null;for(int i = 0 ; i < n;i++){ListNode newNode = new ListNode(sc.nextInt());if(head==null){head = newNode;tail = newNode;}else{tail.next = newNode;tail = newNode;}}System.out.println("输入第一个元素");int left = sc.nextInt();System.out.println("输入第二个元素");int right = sc.nextInt();ListNode forRes = reverseList(head,left,right);while(forRes!=null){System.out.print(forRes.val+" ");forRes = forRes.next;}}
}