LEETCODE: Reverse Linked List II

最新推荐文章于 2025-11-25 00:06:26 发布

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本文介绍了一种在单次遍历中反转链表指定区间的算法实现。从位置m到n，该方法能在原地完成链表的反转操作，并提供了一个具体的C++实现示例。

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.

今天写代码的状态真是不好！这道题居然写了很久。不过打完收工，睡觉去了。

/**
 * 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(m == n)  return head;
        
        // If m == 1, means that a new head should be return;
        ListNode *newhead = NULL;
        // Find the last one before m
        ListNode *prereverse = NULL;
        for(int ii = 0; ii < m - 1; ii ++) {
            if(ii == 0) {
                prereverse = head;
            }
            else {
                prereverse = prereverse->next;
            }
        }
        
        // Reverse nodes from m to n.
        ListNode *newshortend = m == 1 ? head : prereverse->next;
        ListNode *start = m == 1 ? head : prereverse->next;
        ListNode *end = NULL;
		ListNode *current = start->next;
        for(int ii = m; ii < n; ii ++) {
			end = current->next;
			ListNode *temp = current->next;
            current->next = start;
            start = current;
			current = temp;
        }
        
        if(m == 1) {
            newhead = start;
            newshortend->next = current;
            return newhead;
        }
        else {
            prereverse->next = start;
            newshortend->next = end;
            return head;
        }
    }
};