leetcode_ReverseLinkedListII

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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.

很多边界问题很难在短时间写出bug free的代码

/** * 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) {        ListNode *ho=NULL, *p=NULL,*s=NULL, *h=NULL;        ListNode *newhead = new ListNode(-1);        p=newhead;        p->next = head;        for(int i =1 ;i<=m-1;i++)        {            p=p->next;        }        ho=h=p->next;    //保存这个节点        p->next=NULL;        for(int i =m; i<=n ; i++){            s= h;            h=h->next;            s->next=p->next;            p->next = s;        }        ho->next = h;        return newhead->next;            }};

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