vivo 部分链表反转

方法一：使用栈交换需要反转的数字

#include <iostream>
#include <stack>
#include "list.h"
using namespace std;

void partreverse(ListNode* phead, int m, int n)
{
	int count = 1;
	ListNode* p = phead;
	stack<ListNode*> s1;
	if (m < 1 || n < 1)
		return;
	if (m > n)
	{//保证m小于n
		int temp = m;
		m = n;
		n = temp;
	}
	if (m == n)//两者相等不交换直接返回
	　　return;
	while (count < m && p != nullptr)
	{
		count++;
		p = p->m_pNext;
	}
	ListNode* start = p;
	int estart = m + (n - m) / 2 + 1;
	while (count <n  && p != nullptr)
	{
		count++;
		p = p->m_pNext;
		if (count >= estart)
		{
			if (p != nullptr)
				s1.push(p);//后半部分进栈
			else
				return;//超出长度直接返回
		}
	}
	while (!s1.empty())
	{//依次交换
		ListNode* cur = s1.top();
		int temp = cur->m_nValue;
		cur->m_nValue = start->m_nValue;
		start->m_nValue = temp;
		s1.pop();
		start = start->m_pNext;
	}
	return;
}

void Test1()
{
	ListNode* pNode1 = CreateListNode(1);
	ListNode* pNode2 = CreateListNode(2);
	ListNode* pNode3 = CreateListNode(3);
	ListNode* pNode4 = CreateListNode(4);
	ListNode* pNode5 = CreateListNode(5);

	ConnectListNodes(pNode1, pNode2);
	ConnectListNodes(pNode2, pNode3);
	ConnectListNodes(pNode3, pNode4);
	ConnectListNodes(pNode4, pNode5);
	
	PrintList(pNode1);
	partreverse(pNode1, 2, 4);
	PrintList(pNode1);
}

int main(int argc, char* argv[])
{
	Test1();
	return 0;
}

方法二：直接反转需要反转的部分

#include <iostream>
#include "list.h"
using namespace std;

void partreverse(ListNode* phead, int m, int n)
{
    int count = 1;
    ListNode* p = phead;
    if (m < 1 || n < 1)
        return;
    if (m > n)
    {//保证m小于n
        int temp = m;
        m = n;
        n = temp;
    }
　　if (m == n)//两者相等不交换直接返回
        return;
    while (count < m-1 && p != nullptr)
    {
        count++;
        p = p->m_pNext;
    }
    ListNode* prev = p, *start = phead, *end = nullptr, *next = nullptr;
    if (m == 1)
        prev = nullptr;
    if (prev != nullptr)
        start = prev->m_pNext;
    p = start;
    for (count = m; count < n && p != nullptr;)
    {
        count++;
        p = p->m_pNext;
    }
    end = p;
    if (end != nullptr)
        next = end->m_pNext;
    else
        return;//end为nullptr，超出链表长度
    if(prev !=nullptr)
        prev->m_pNext = end;//prev指向反转后的链表头
    end->m_pNext = nullptr;
    ListNode* rprve = start, *rcur = start->m_pNext, *rnext = rcur->m_pNext;
    while (rcur != end) 
    {//指定范围内的链表反转
        rcur->m_pNext = rprve;
        rprve = rcur;
        rcur = rnext;
        if (rnext != nullptr)
            rnext = rnext->m_pNext;
    }
    rcur->m_pNext = rprve;
    start->m_pNext = next;
    if (prev == nullptr)
        phead = end;//链表头被反转
    PrintList(phead);
    return;

}

void Test1()
{
    ListNode* pNode1 = CreateListNode(1);
    ListNode* pNode2 = CreateListNode(2);
    ListNode* pNode3 = CreateListNode(3);
    ListNode* pNode4 = CreateListNode(4);
    ListNode* pNode5 = CreateListNode(5);

    ConnectListNodes(pNode1, pNode2);
    ConnectListNodes(pNode2, pNode3);
    ConnectListNodes(pNode3, pNode4);
    ConnectListNodes(pNode4, pNode5);
    
    PrintList(pNode1);
    partreverse(pNode1, 1, 5);
    //PrintList(pNode1);
}

int main(int argc, char* argv[])
{
    Test1();
    return 0;
}

前面花太多时间导致后面的送分题01背包没时间做，我恨！