LeetCode OJ：Reverse Linked List II（反转链表II）

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.

大清早起来就被链表虐哭了啊， 看了下别人的，额原来可以这么简单，果然脑子还是转不过来的，实际上是很常见的一个题目，代码很简单，完全不用注释也可以的：

/**
 * 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) {
        if(head == NULL) return NULL;
        ListNode * p = head;
        int i, j;
        for(i = 1; i < m; ++i){
            p = p->next;
        }
        ListNode * q = p;
        for(i = m; i < n; ++i){
            for(j = i; j < n; ++j){
                q = q->next;
            }
            swap(p->val, q->val);
            n--;
            p = p->next;
            q = p;
        }
        return head;
    }
};

注意一下那个swap， swap用的很巧妙。

之后有看到一个大神写出来的，也很简单，贴出来学习一个：

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (head == NULL)
            return NULL;
            
        ListNode *q = NULL;
        ListNode *p = head;
        for(int i = 0; i < m - 1; i++)
        {
            q = p;
            p = p->next;
        }
        
        ListNode *end = p;
        ListNode *pPre = p;
        p = p->next;
        for(int i = m + 1; i <= n; i++)
        {
            ListNode *pNext = p->next;
            
            p->next = pPre;
            pPre = p;
            p = pNext;
        }
        
        end->next = p;
        if (q)
            q->next = pPre;
        else
            head = pPre;
        
        return head;
    }
};

唉唉，经常遇到链表脑子就转不过来，这个还是要多练练啊。

下面贴一个java版本的,方法基本上和第一种差不多,熟悉一下用法：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
public class Solution {
    public ListNode reverseBetween(ListNode head, int m, int n) {
        int count = 1;
        ListNode helper = new ListNode(-1);
        helper.next = head;
        ListNode p = helper.next;
        ListNode pPre = helper;
        while(count != m){
            p = p.next;
            pPre = pPre.next;
            count++;
        }
        ListNode midPre = pPre;//第一个节点的位置
        ListNode tmp = null;
        while(count != n){
            tmp = p.next;
            p.next = pPre;
            pPre = p;
            p = tmp;
            count++;
        }
//        ListNode mid2= p; // 指向第二个节点的位置
        tmp = p.next;
        p.next = pPre;
        midPre.next.next = tmp;
        midPre.next = p;
        return helper.next;
    }
}