【LeetCode】Reverse Linked List II

標籤：leetcode   鏈表   

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.

思路：局部反轉鏈表，然後在合并。

/** * 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)        {            return head;        }        ListNode *sg = new ListNode(-1);        sg->next = head;        head = sg;        ListNode *pre = head;        for(int i = 1; i < m; i++)        {            pre = pre->next;        }        ListNode *cur = pre->next;        ListNode *cur_next = cur->next;        if(n - m > 0)        {            int step = n - m;            while(step > 0 && cur_next != NULL)            {                ListNode *temp = cur_next->next;                cur_next->next = cur;                cur = cur_next;                cur_next = temp;                step--;            }            ListNode *temp = pre->next;            pre->next = cur;            temp->next = cur_next;        }        head = head->next;        delete sg;        return head;    }};