92、Reverse Linked List II

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

题目要求只能遍历一次。

首先需要找到第一个开始变换结点的前一个节点（往前走m-1步），然后开始交换，每次交换两个点，开始变换位置的节点和后一个节点（即 cur->next 节点和 pre->next节点），需要n-m步即可。

c++11

/**
 * 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) {
        //新建一个节点是怕m=1的时候，当前节点没有前一个节点可以操作
        ListNode* res = new ListNode(-1);
        res->next = head;
        ListNode* pre = res;
        //找到开始变换的前一个节点
        for (int i = 0; i < m-1; i++) pre = pre->next;
        ListNode* cur = pre->next;//开始变换节点
        for (int i = m; i < n; i++) {
            ListNode* tmp = cur->next;//保存开始变换节点的下一个节点
            cur->next = tmp->next;
            tmp->next = pre->next;
            pre->next = tmp;
        }
        return res->next;
    }
};

ac结果

untime: 4 ms, faster than 100.00% of C++ online submissions for Reverse Linked List II.
Memory Usage: 8.9 MB, less than 41.95% of C++ online submissions for Reverse Linked List II.