LeetCode 92 Reverse Linked List II（翻转链表II）（Linked List）（*）

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;
    }
}