翻转链表

翻转链表中第m个节点到第n个节点的部分

样例

例1:

输入: 1->2->3->4->5->NULL, m = 2 and n = 4,
输出: 1->4->3->2->5->NULL.

例2:

输入: 1->2->3->4->NULL, m = 2 and n = 3,
输出: 1->3->2->4->NULL.

挑战

Reverse it in-place and in one-pass

注意事项
m，n满足1 ≤ m ≤ n ≤ 链表长度

源码

/**
 * Definition of singly-linked-list:
 * class ListNode {
 * public:
 *     int val;
 *     ListNode *next;
 *     ListNode(int val) {
 *        this->val = val;
 *        this->next = NULL;
 *     }
 * }
 */

class Solution {
public:
    /**
     * @param head: ListNode head is the head of the linked list 
     * @param m: An integer
     * @param n: An integer
     * @return: The head of the reversed ListNode
     */
    ListNode * reverseBetween(ListNode * head, int m, int n) {
        // write your code here
        ListNode * p1,*psre,*pere,*ps,*pe,*penext;
        
        if(m==n)return head;
        
        p1=head;
        pe=ps=p1;
        if(p1->next!=NULL)
            for(int i=1;p1!=NULL;i++,p1=p1->next){
                if(i==m){
                    ps=p1;
                }
                if(i==n){
                    pe=p1;
                }
                if(i<m){
                    psre=p1;
                }
                if(i<n){
                    pere=p1;
                }
            }
        
        // if(p1->next!=NULL)
           
        penext=pe->next;
            
        if(m!=1)
            psre->next=pe;
        else{
            head=pe;
        }
        
        ListNode *tp,*tpnext,*temp;
        tp=ps;
        if(head->next!=NULL){
        tpnext=ps->next;
        ps->next=penext;
        }
        
        if(n-m!=0){

            for(int i=0;i<n-m;i++){
                temp=tpnext->next;
                tpnext->next=tp;
                tp=tpnext;
                tpnext=temp;
            }
        }
        return head;
    }
};