LeetCode——Reverse Linked List II

最新推荐文章于 2025-12-21 17:04:13 发布

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本文详细阐述了如何在不使用额外空间的情况下，一次遍历链表，实现从位置m到n的区间反转。

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.

» Solve this problem

细心一点，直接挪。

关键点有四个，1.第m-1个点，2.第m个点，3.第n个点，4.第n+1个点。

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        ListNode *before, *start, *end, *after;
        before = after = start = end = NULL;        
        ListNode *ptr = head;
        for (int i = 1; i < m; i++) {
            before = ptr;
            ptr = ptr->next;            
        }
        start = end = ptr;
        for (int i = m; i <= n; i++) {
            after = ptr->next;
            ptr->next = end;
            end = ptr;
            ptr = after;            
        }
        if (before == NULL) {
            head = end;
        }
        else {
            before->next = end;
        }
        start->next = after;
        return head;
    }
};