LeetCode之“链表”：Reverse Linked List && Reverse Linked List II

　 1. Reverse Linked List

　　题目链接

　　题目要求：

　　Reverse a singly linked list.

　　Hint:

　　A linked list can be reversed either iteratively or recursively. Could you implement both?

　　初想用递归来实现应该会挺好的，但最终运行时间有点久，达到72ms，虽然没超时。具体程序如下：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */

ListNode* recurList(ListNode* p)
{
    if (p->next != NULL)
    {
        ListNode* tmp = p->next;
        if (tmp->next == NULL)
        {
            tmp->next = p;
            p->next = NULL;
            return tmp;
        }
    }

    ListNode* ret = recurList(p->next);
    ListNode* tmp = ret;
    while (tmp->next != NULL)
        tmp = tmp->next;
    tmp->next = p;
    p->next = NULL;
    return ret;
}

ListNode* reverseList(ListNode* head) {
    if (head == NULL || head->next == NULL)
        return head;

    ListNode* tail = recurList(head);
    return tail;
}

　　在LeetCode论坛发现了一个8ms的解法，不得不为其简洁高效合彩！具体程序如下：

ListNode* reverseList(ListNode* head) {
    ListNode *curr = head, *prev = nullptr;
    while (curr) {
        auto next = curr->next;
        curr->next = prev;
        prev = curr, curr = next;
    }
    return prev;
}

 　 2. Reverse Linked List II

　　题目链接

　　题目要求：

　　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.

　　该题的解法可以将一个链表分成三个部分，即1~m-1、m~n、n+1~最后一个元素3个部分，且仅对中间部分进行翻转，最后再将这3个部分合并。具体程序如下：

/**
 * 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 || !head->next || m >= n)
             return head;
        
        ListNode *start = head;
        ListNode *preList = nullptr, *postList = nullptr;
        
        int k = 1;
        while(k < m)
        {
            preList = start;
            start = start->next;
            k++;
        }
        
        ListNode *prev = nullptr;
        while(k <= n)
        {
            if(k == n)
                postList = start->next;
            auto next = start->next;
            start->next = prev;
            prev = start;
            start = next;
            k++;
        }
        
        ListNode *end = prev;
        while(end && end->next)
            end = end->next;
        
        if(preList)
            preList->next = prev;
        else
            head = prev;
        end->next = postList;
        
        return head;
    }
};