给定一个链表,翻转该链表从m到n的位置。要求直接翻转而非申请新空间。
#include <iostream>
#include <vector>
#include <algorithm>
#include <map>
#include <cstdio>
using namespace std;
typedef struct tagSNode
{
int value;
tagSNode* pNext;
tagSNode(int v) :value(v), pNext(NULL){}
}SNode;
void Reverse(SNode* pHead, int from, int to)
{
SNode* pCur = pHead->pNext;
int i;
for (i = 0; i<from - 1; i++)
{
pHead = pCur;
pCur = pCur->pNext;
}
SNode* pPre = pCur;
pCur = pCur->pNext;
to--;
SNode* pNext;
for (; i<to; i++)
{
//此处需要结合图示
pNext = pCur->pNext;
pCur->pNext = pHead->pNext;
pHead->pNext = pCur;
pPre->pNext = pNext;
pCur = pNext;
}
}
void Destroy(SNode* p)
{
SNode* next;
while (p)
{
next = p->pNext;
delete p;
p = next;
}
}
int main()
{
SNode* pHead = new SNode(0);
int i;
for (i = 0; i<10; i++)
{
SNode* p = new SNode(rand() % 100);
p->pNext = pHead->pNext;
pHead->pNext = p;
}
//Print(pHead);
Reverse(pHead, 4, 8);
//Print(pHead);
//Print(pHead);
Destroy(pHead);
return 0;
}