[leetcode 92] Reverse Linked List II

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#链表

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本文介绍了一种在链表中反转指定区间的节点的方法，包括实现细节和代码示例。

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.

/**
 * 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 || head->next == NULL || (n-m < 1)) {
            return head;
        }
        ListNode dummy(-1);
        ListNode *prev = &dummy;
        dummy.next = head;
        auto cur = head;
        for (int i = 0; i < m - 1; i++) {
            prev = prev->next;
            cur = cur->next;
        }
        for (int i = m; i < n; i++) {
            cur = cur->next;
        }
        prev->next = reverse(prev, prev->next, cur);
        return dummy.next;
        
    }
    ListNode *reverse(ListNode *prev, ListNode *begin, ListNode *end) {
        ListNode *end_next = end->next;
        ListNode *p = begin;
        ListNode *cur = p->next;
        ListNode *next = cur?cur->next:NULL;
        for ( ;cur != end_next; p = cur, cur = next, next = next?next->next:NULL) {
            cur->next = p;
        }
        begin->next = end_next;
        return end;
    }
};