92. Reverse Linked List II

Problem:

Reverse a linked list from position m to n. Do it in one-pass.

Note: 1 ≤ m ≤ n ≤ length of list.

Example:

Input: 1->2->3->4->5->NULL, m = 2, n = 4
Output: 1->4->3->2->5->NULL

思路：

保存2个节点pre和cur，其中pre指向第m-1个节点（从1开始算），而cur指向第m个节点。在每次循环中，cur->next指向的节点插入到pre和pre->next指向的节点中，然后经过操作cur->next = cur->next->next后cur->next指向的节点右移一位，经过n-m次操作后cur->next指向第n+1个节点，然后经过n-m次插入操作后原本m后面的节点m+1到n节点倒序排列在m-1和m（现在的第n节点）节点之间。

Solution (C++):

ListNode* reverseBetween(ListNode* head, int m, int n) {
    ListNode *dummy = new ListNode(0), *pre = dummy, *cur;
    dummy->next = head;
    for (int i = 0; i < m - 1; ++i) {
        pre = pre->next;
    }
    cur = pre->next;    //pre指向第m-1个节点，cur指向第m个节点
    for (int i = 0; i < n - m; ++i) {       //每次循环时把第m+i+1个节点插入到（重新排列之后的）第m-1个节点和第m个节点之间
        ListNode *tmp = pre->next;
        pre->next = cur->next;
        cur->next = cur->next->next;
        pre->next->next = tmp;
    }
    return dummy->next;
}

性能：

Runtime: 4 ms  Memory Usage: 8.9 MB