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Reverse Linked List II
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.
C++实现代码：
#include<iostream>#include<new>using namespace std;//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) return NULL; ListNode *p=head; ListNode *pre=head; ListNode *qpre=NULL; ListNode *q=NULL; int count=0; while(p) { count++; if(count==m) { qpre=pre; q=p; } else if(count==n) break; pre=p; p=p->next; } cout<<p->val<<endl; cout<<q->val<<endl; if(p==NULL||p==q) return head; ListNode *pp=p->next; ListNode *qq=NULL; p->next=NULL; p=q->next; cout<<p->val<<endl; q->next=pp; while(p) { qq=p->next; cout<<p->val<<endl; p->next=q; q=p; cout<<q->val<<endl; p=qq; } cout<<q->val<<endl; if(m!=1) qpre->next=q; else head=q; return head; } void createList(ListNode *&head) { ListNode *p=NULL; int i=0; int arr[10]= {9,8,5,4,4,3,3,3,2,1}; for(i=0; i<3; i++) { if(head==NULL) { head=new ListNode(arr[i]); if(head==NULL) return; } else { p=new ListNode(arr[i]); p->next=head; head=p; } } }};int main(){ Solution s; ListNode *L=NULL; s.createList(L); ListNode *head=L; while(head) { cout<<head->val<<" "; head=head->next; } cout<<endl; L=s.reverseBetween(L,2,3); while(L) { cout<<L->val<<" "; L=L->next; }}
Reverse Linked List II
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