我们接下来要做的是,先把Mpre的next域指向NodeM节点的后一个节点; 再把NodeM所在节点移动到NodeN所在节点之后,使得NodeN的next域指向NodeM所在节点,NodeM所在节点next域指向NodeN的next域所指节点; 然后让NodeM指向Mpre的next域指向的节点; 演示图2 然后再重复上面的步骤; 这是NodeM和NodeN相遇,反转完成。 代码...
https://oj.leetcode.com/problems/reverse-linked-list-ii/ Reverse a linked list from positionmton. Do it in-place and in one-pass. For example: Given1->2->3->4->5->NULL,m= 2 andn= 4, return1->4->3->2->5->NULL. Note: Givenm,nsatisfy the following condition: 1≤m≤n≤ ...
Reverse Linked List II Reverse a linked list from positionmton. Do it in-place and in one-pass. For example: Given1->2->3->4->5->NULL,m= 2 andn= 4, return1->4->3->2->5->NULL. Note: Givenm,nsatisfy the following condition: 1≤m≤n≤ length of list. 把[m,n]那一段抠...
Given the head of a singly linked list and two integers left and right where left <= right, reverse the nodes of the list from position left to position right, and return the reversed list. Example 1: Input: head = [1,2,3,4,5], left = 2, right = 4 Output: [1,4,3,2,5] ...
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example: Given 1->2->3->4->5->NULL, m = 2 and n = 4, return 1->4->3->2->5->NULL. Note: Given m, n satisfy the following condition: ...
Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list. Example: Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL 1. 2. 题目大意 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...
publicListNodereverseBetween(ListNodehead,intm,intn){if(m==n){returnhead;}ListNodedummy=newListNode(0);dummy.next=head;intcount=0;ListNodeleft1=null;ListNodeleft2=null;ListNodepre=dummy;while(head!=null){count++;//到达 m,保存 l1 和 l2if(count==m){left1=pre;left2=head;}// m 和 n ...
/** * Definition for singly-linked list. * type ListNode struct { * Val int * Next *ListNode * } */ func reverseBetween(head *ListNode, left int, right int) *ListNode { vhead:=&ListNode{} vhead.Next = head var f, l, r, t *ListNode f=vhead for i:=1;i<left;i++ { f=...
Reverse a linked list from position m to n. Do it in one-pass. Example: Input: 1->2->3->4->5->NULL, m = 2, n = 4 Output: 1->4->3->2->5->NULL Note: 1≤ m ≤ n ≤ length of list. 解释下题目: 在一次循环中把链表指定位置倒序 ...
return new ImageLayers<>(ImmutableList.copyOf(layers), layerDigestsBuilder.build()); } // LinkedHashSet maintains the order but keeps the first occurrence. Keep last occurrence by // adding elements in reverse, and then reversing the result Set<T> dedupedButReversed = new LinkedHashSet<T>...