Reverse Linked List II

Q:

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.

A:

其实还是比较简单的，但是凭想象去交换元素的时候还是让我很烦躁，空间思维能力极差无比。。。。。。。。

没有去考虑n<m的情况，题目已经告知不用考虑这个，如果考虑的话还需要先数链表长度，当然这些问题都不重要，重要的是你需要考虑到并确认，OK

 

/**
 * 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 (m == n) return head;
        if (m > n) swap(m, n);
        ListNode* pre = NULL;
        ListNode* cur = head;
        ListNode* sub_head = NULL;
        for (int i = 1; i < m; ++i) {
            pre = cur;
            cur = cur->next;
        }
        sub_head = cur;
        for (int i = m; i < n; ++i) {
            ListNode* tmp1 = cur->next;
            ListNode* tmp2 = cur->next->next;
            cur->next->next = sub_head;
            cur->next = tmp2;
            sub_head = tmp1;
        }
        if (!pre) {
            head = sub_head;
        } else {
            pre->next = sub_head;
        }
        return head;
    }
};