92. Reverse Linked List II【Medium】
92. Reverse Linked List II【Medium】
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:
1 ≤ m ≤ n ≤ length of list.
解法一:
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
if (head == NULL || head->next == NULL) {
return head;
} ListNode * dummy = new ListNode(INT_MIN);
dummy->next = head;
ListNode * mth_prev = findkth(dummy, m - );
ListNode * mth = mth_prev->next;
ListNode * nth = findkth(dummy, n);
ListNode * nth_next = nth->next;
nth->next = NULL; reverseList(mth); mth_prev->next = nth;
mth->next = nth_next; return dummy->next;
} ListNode *findkth(ListNode *head, int k)
{
for (int i = ; i < k; i++) {
if (head == NULL) {
return NULL;
}
head = head->next;
}
return head;
} ListNode * reverseList(ListNode * head)
{
ListNode * pre = NULL;
while (head != NULL) {
ListNode * next = head->next;
head->next = pre;
pre = head;
head = next;
} return pre;
}
};
解法二:
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
if (head == NULL || head->next == NULL) {
return head;
} ListNode * dummy = new ListNode(INT_MIN);
dummy->next = head;
head = dummy; for (int i = ; i < m; ++i) {
if (head == NULL) {
return NULL;
}
head = head->next;
} ListNode * premNode = head;
ListNode * mNode = head->next;
ListNode * nNode = mNode;
ListNode * postnNode = mNode->next; for (int i = m; i < n; ++i) {
if (postnNode == NULL) {
return NULL;
}
ListNode * temp = postnNode->next;
postnNode->next = nNode;
nNode = postnNode;
postnNode = temp;
} mNode->next = postnNode;
premNode->next = nNode; return dummy->next;
}
};
解法三:
public ListNode reverseBetween(ListNode head, int m, int n) {
if(head == null) return null;
ListNode dummy = new ListNode(); // create a dummy node to mark the head of this list
dummy.next = head;
ListNode pre = dummy; // make a pointer pre as a marker for the node before reversing
for(int i = ; i<m-; i++) pre = pre.next; ListNode start = pre.next; // a pointer to the beginning of a sub-list that will be reversed
ListNode then = start.next; // a pointer to a node that will be reversed // 1 - 2 -3 - 4 - 5 ; m=2; n =4 ---> pre = 1, start = 2, then = 3
// dummy-> 1 -> 2 -> 3 -> 4 -> 5 for(int i=; i<n-m; i++)
{
start.next = then.next;
then.next = pre.next;
pre.next = then;
then = start.next;
} // first reversing : dummy->1 - 3 - 2 - 4 - 5; pre = 1, start = 2, then = 4
// second reversing: dummy->1 - 4 - 3 - 2 - 5; pre = 1, start = 2, then = 5 (finish) return dummy.next; }
参考了@ardyadipta 的代码,Simply just reverse the list along the way using 4 pointers: dummy, pre, start, then
92. Reverse Linked List II【Medium】的更多相关文章
- 82. Remove Duplicates from Sorted List II【Medium】
82. Remove Duplicates from Sorted List II[Medium] Given a sorted linked list, delete all nodes that ...
- 92. Reverse Linked List II
题目: Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1- ...
- 【LeetCode】92. Reverse Linked List II 解题报告(Python&C++)
作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 迭代 递归 日期 题目地址:https://leet ...
- [LeetCode] 92. Reverse Linked List II 倒置链表之二
Reverse a linked list from position m to n. Do it in one-pass. Note: 1 ≤ m ≤ n ≤ length of list. Exa ...
- [LeetCode] 92. Reverse Linked List II 反向链表II
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1-> ...
- 【leetcode】92. Reverse Linked List II
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1-> ...
- 【一天一道LeetCode】#92. Reverse Linked List II
一天一道LeetCode 本系列文章已全部上传至我的github,地址:ZeeCoder's Github 欢迎大家关注我的新浪微博,我的新浪微博 欢迎转载,转载请注明出处 (一)题目 Reverse ...
- 【Leetcode】92. Reverse Linked List II && 206. Reverse Linked List
The task is reversing a list in range m to n(92) or a whole list(206). All in one : U need three poi ...
- leetcode 92 Reverse Linked List II ----- java
Reverse a linked list from position m to n. Do it in-place and in one-pass. For example:Given 1-> ...
随机推荐
- luogu P1418 选点问题
题目描述 给出n个点,m条边,每个点能控制与其相连的所有的边,要求选出一些点,使得这些点能控制所有的边,并且点数最少.同时,任意一条边不能被两个点控制 输入输出格式 输入格式: 第一行给出两个正整数n ...
- [Codeforces 8E] Beads
Brief Intro: 将所有n位二进制串中满足字典序不大于其逆序串,取反串,逆序取反串中按字典序排序的第m个输出 n<=50 Algorithm: 首次接触数位DP的题目 根据数据范围,我们 ...
- URAL 1349 Farm
Discription Here is a farm. Here is a farmer that counts how many animal live in his farm: a camels, ...
- [Contest20180116]随机游走
题意:给一棵树,多次询问$a$到$b$期望步数,每一步都是随机的 对期望DP了解更深入了一些 先预处理$up_x$表示从$x$走到$fa_x$的期望步数 可以直接往上走,也可以先去儿子再回来,设$x$ ...
- [CF911C]Three Garlands
题目大意: 给你三个灯,分别以k1秒一次,k2秒一次和k3秒一次的频率闪烁着. 你可以自定义三个灯开启的时间,问是否有一种方案,使得max(k1,k2,k3)秒之后,每秒钟都至少有一盏灯闪烁. 思路: ...
- iOS 获取自定义cell上按钮所对应cell的indexPath.row的方法
在UITableView或UICollectionView的自定义cell中创建一button,在点击该按钮时知道该按钮所在的cell在UITableView或UICollectionView中的行数 ...
- 【java JVM】JVM中类的加载,加载class文件的原理机制
暂时贴图一张,以后补充 解释: 1.符号引用替换为直接引用[参考:http://blog.csdn.net/maerdym/article/details/8087620] 在java中,一个java ...
- Android线程与线程池
引言 在Android中,几乎完全采用了Java中的线程机制.线程是最小的调度单位,在很多情况下为了使APP更加流程地运行,我们不可能将很多事情都放在主线程上执行,这样会造成严重卡顿(ANR),那么这 ...
- Go语言的9大优势和3大缺点, GO语言最初的定位就是互联网时代的C语言, 我为什么放弃Go语言
Go语言的9大优势和3大缺点 转用一门新语言通常是一项大决策,尤其是当你的团队成员中只有一个使用过它时.今年 Stream 团队的主要编程语言从 Python 转向了 Go.本文解释了其背后的九大原因 ...
- web UIproject师必读的一篇文档,写的很好
今天在花瓣网上看到的一片文章.写的是web UIproject师的工作应该是如何的,内容非常精彩,分享给小伙伴.希望web UIproject师看了之后有所启示. 我不是一个简单的web UI htt ...