LeetCode 92 Reverse Linked List II（翻转链表II）（Linked List）（*）
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翻译
将一个链表中位置m和n的节点进行翻转。就地且一次通过。
例如
给定 1->2->3->4->5->NULL, m = 2 和n = 4,
返回 1->4->3->2->5->NULL.
备注：
给定的m和n满足以下条件：
1 <= m <= n <= 链表的长度
原文
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.分析
C Plus Plus
/**
* Definition for singly-linked list.
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
class Solution {
public:
ListNode* reverseBetween(ListNode* head, int m, int n) {
ListNode *newHead = new ListNode(0);
newHead->next = head;
ListNode *c = newHead;
for (int i = 0; i < m - 1; i++) {
c = c->next;
}
ListNode *p = c->next;
for (int i = 0; i < n - m; i++) {
ListNode *t = p->next;
p->next = t->next;
t->next = c->next;
c->next = t;
}
return newHead->next;
}
};Java
updated at 2016/09/23
/**
* Definition for singly-linked list.
* public class ListNode {
* int val;
* ListNode next;
* ListNode(int x) { val = x; }
* }
*/
public class Solution {
ListNode reverseBetween(ListNode head, int m, int n) {
ListNode newHead = new ListNode(0);
newHead.next = head;
ListNode c = newHead;
for (int i = 0; i < m - 1; i++) {
c = c.next;
}
ListNode p = c.next;
for (int i = 0; i < n - m; i++) {
ListNode tmp = p.next;
p.next = tmp.next;
tmp.next = c.next;
c.next = tmp;
}
return newHead.next;
}
}