3.给定一个链表，两个整数m，n，m<n，翻转链表中m到n的元素

#include <iostream>
using namespace std;

typedef struct tagSNode{
	int value;
	tagSNode* pNext;
	tagSNode(int v) : value(v), pNext(NULL){}
}SNode;

void Print(SNode *p){
	SNode *q = p;
	while (q != NULL){
		cout << q->value<<" ";
		q = q->pNext;
	}
	cout << endl;
}

void Destory(SNode *p){
	SNode *next;
	while (p){
		next = p->pNext;
		delete p;
		p = next;
	}
}

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++){
		pNext = pCur->pNext;
		pCur->pNext = pHead->pNext;
		pHead->pNext = pCur;
		pPre->pNext = pNext;
		pCur = pNext;
	}
}

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);
	Destory(pHead);
	return 0;
}

4、给定一个链表和一个值x，将链表划分成两部分，使得划分后小于x的结点在前，大于等于x的结点在后，并保持原链表中的顺序

void Partition(SNode *pHead, int pivotKey){
	SNode *pLeftHead = new SNode(0);
	SNode *pRightHead = new SNode(0);

	//两个链表的当前最后一个元素
	SNode *left = pLeftHead;
	SNode *right = pRightHead;
	SNode *p = pHead->pNext;
	while (p){
		if (p->value < pivotKey){
			left->pNext = p;
			left = p;
		}
		else{
			right->pNext = p;
			right = p;
		}
		p = p->pNext;
	}

	left->pNext = pRightHead->pNext;
	right->pNext = NULL;

	pHead->pNext = pLeftHead->pNext;

	delete pLeftHead;
	delete pRightHead;
}