标签：nec ptr ace col 数字 style 时间 normal 指定
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！
标签：nec ptr ace col 数字 style 时间 normal 指定
原文地址：https://www.cnblogs.com/ccxikka/p/10976784.html
