92. Reverse Linked List II

最新推荐文章于 2024-01-12 14:53:46 发布

无名_1989

最新推荐文章于 2024-01-12 14:53:46 发布

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本文介绍了一种在单次遍历中就地反转链表中从位置 m 到 n 的算法实现。通过 C++ 示例代码展示了如何高效地完成这一任务，并确保了边界条件的正确处理。

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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.

class Solution {
public:
    ListNode* reverseBetween(ListNode* head, int m, int n) {
        if(head == NULL || head->next == NULL || m == n) return head;
        if(n<m) return NULL;
        ListNode * p = head;
        ListNode * tmp_head = new ListNode(0);
        tmp_head->next = p;
        ListNode * pre = tmp_head;
        
        int count = m - 1;
        
        while(count&&p)
        {
            pre = p;
            p = p->next;
            --count;
        }
        
        if (p == NULL) return NULL;
        
        count = n-m;
        
        
        ListNode * next_node = p->next;
        ListNode * first_reverse_node = p ;
        while(count&&next_node)
        {
            cout<<next_node->val<<"##"<<endl;
            ListNode * tmp = pre->next;
            ListNode * next = next_node->next;
            pre->next = next_node;
            first_reverse_node->next = next;
            next_node->next = tmp;
            next_node = next;
            --count;
        }
        
        return tmp_head->next;
    }
};