LeetCode Reverse Linked List II

class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if (head == NULL || n <= m) return head;
        int offset = n - m;

        ListNode* cur = head;
        ListNode* pre = NULL;

        while (cur != NULL && --m > 0) {
            pre = cur;
            cur = cur->next;
        }
        ListNode* start = cur;

        while (cur != NULL && offset-- > 0) {
            cur = cur->next;
        }
        ListNode* end = cur;
        ListNode* end_next = end->next;
        end->next = NULL;
        ListNode* rhead = reverse(start);

        if (pre == NULL) {
            head = rhead;
        } else {
            pre->next = rhead;
        }
        start->next = end_next;
        return head;
    }

    ListNode* reverse(ListNode* head) {
        ListNode* cur = head;
        ListNode* pre = NULL;
        while (cur != NULL) {
            ListNode* t = cur->next;
            cur->next = pre;
            pre = cur;
            cur = t;
        }
        return pre;
    }
};

 链表指针还是要小心着点，写着写着把前面想的要注意的地方给漏了

第二轮：

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.

第一次做的时候应该不算是one-pass

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
 // 10:55
class Solution {
public:
    ListNode *reverseBetween(ListNode *head, int m, int n) {
        if (head == NULL || head->next == NULL) {
            return head;
        }
        ListNode fakeHead(0);
        fakeHead.next = head;
        
        int k = m;
        ListNode* pre = NULL;
        ListNode* cur = &fakeHead;
        while (k--) {
            pre = cur;
            cur = cur->next;
        }
        ListNode* first_tail = pre;
        ListNode* reversed_tail = cur;
        
        pre = NULL;
        k = n - m + 1;
        while (k--) {
            ListNode* tmp = cur->next;
            cur->next = pre;
            pre = cur;
            cur = tmp;
        }
        ListNode* reversed_head = pre;
        ListNode* second_head = cur;
        
        first_tail->next = reversed_head;
        reversed_tail->next = second_head;
        return fakeHead.next;
    }
};