版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/u011729265/article/details/12209101
1.链表翻转。给出一个链表和一个数k，比如链表1→2→3→4→5→6，k=2，则翻转后2→1→4→3→6→5，若k=3,翻转后3→2→1→6→5→4，若k=4，翻转后4→3→2→1→5→6，用程序实现。
#include <iostream>
using namespace std;
struct ListNode
{
int m_nValue;
ListNode *m_pNext;
};
ListNode* CreateList(int val)
{
ListNode *pHead = new ListNode;
pHead->m_nValue = val;
pHead->m_pNext = NULL;
return pHead;
}
void InsertNode(ListNode **pHead, int val)
{
ListNode *pNode = new ListNode;
pNode->m_nValue = val;
pNode->m_pNext = NULL;
while ((*pHead)->m_pNext != NULL)
{
(*pHead) = (*pHead)->m_pNext;
}
(*pHead)->m_pNext = pNode;
(*pHead) = pNode;
}
void PrintList(ListNode *pHead)
{
while (pHead != NULL)
{
cout<<pHead->m_nValue<<" ";
pHead = pHead->m_pNext;
}
cout<<endl;
}
ListNode* Reverse(ListNode *pHead)
{
if (pHead == NULL || pHead->m_pNext == NULL)
{
return pHead;
}
ListNode *pPre = NULL;
ListNode *pCurrent = pHead;
ListNode *pPost = pHead->m_pNext;
while (pCurrent->m_pNext != NULL)
{
pCurrent->m_pNext = pPre;
pPre = pCurrent;
pCurrent = pPost;
pPost = pPost->m_pNext;
}
pCurrent->m_pNext = pPre;
return pCurrent;
}
ListNode* ReverseList(ListNode *pHead, int k)
{
if (pHead==NULL || pHead->m_pNext==NULL)
{
return pHead;
}
ListNode *pPre = NULL;
ListNode *pCurrent = pHead;
ListNode *pPost = pHead->m_pNext;
ListNode *pStart = NULL;
ListNode *pEnd = NULL;
int n = 0;
pEnd = pCurrent;
pEnd->m_pNext = NULL;
while (pPost != NULL)
{
++n;
if (n == (k+1))
{
pStart = pPre;
pEnd->m_pNext = ReverseList(pCurrent, k);
return pStart;
}
else
{
pCurrent->m_pNext = pPre;
pPre = pCurrent;
pCurrent = pPost;
pPost = pPost->m_pNext;
}
}
pCurrent->m_pNext = pPre;
pStart = Reverse(pCurrent);
return pStart;
}
int main()
{
ListNode *pHead = NULL;
ListNode *head = NULL;
int n;
cout<<"输入链表中节点的个数 n："<<endl;
cin>>n;
cout<<"请输入n个整数值："<<endl;
for (int i=0; i<n; ++i)
{
int data;
cin>>data;
if (pHead == NULL)
{
pHead = CreateList(data);
head = pHead;
}
else
{
InsertNode(&pHead, data);
}
}
int k;
cout<<"请输入k:"<<endl;
cin>>k;
head = ReverseList(head, k);
PrintList(head);
system("pause");
return 0;
} 2.一个函数access(),调用频率不能超过R次/sec，用程序实现一个函数，当超过R次/sec时返回access false，不超过时返回success
#define false 0
#define success 1
int getcurrentms()
{
struct timeval tv;
gettimeofday(&tv,NULL);
return tv.tv_sec*1000+tv.tv_usec/1000; //得到毫秒数
}
bool count_access()
{
static int count=0;
static int time_ms_old=0,time_ms_now;
if(count==0)
{
time_ms_old=getcurrentms();
}
count++;
access();
if(count>=R)
{
time_ms_now=getcurrentms();
if(time_ms_now-time_ms_pld>=1000)
return false;
else
return success;
}
return success;
}
解: 思路:从矩阵的右上角开始判断即可,每次可以消除一行或一列,详见剑指offer一书.
4.利用两个栈，模拟queue
#include <iostream>
#include <stack>
using namespace std;
template <class T>
class Queue
{
public:
Queue()
{
}
~Queue()
{
}
void add(const T& t);
T remove();
private:
stack<T> s1;
stack<T> s2;
};
template <class T>
void Queue<T>::add(const T& t)
{
s1.push(t);
}
template <class T>
T Queue<T>::remove()
{
if (s2.size() <= 0)
{
while (s1.size() > 0)
{
T t = s1.top();
s2.push(t);
s1.pop();
}
}
if (s2.size() == 0)
{
throw new exception("empty queue");
}
T t = s2.top();
s2.pop();
return t;
}
int main()
{
Queue<char> q;
q.add('A');
q.add('B');
q.add('C');
cout<<q.remove()<<endl;
cout<<q.remove()<<endl;
cout<<q.remove()<<endl;
system("pause");
return 0;
} 5.求两个字符串的最长公共子串
public class MaxConString {
/**
* 计算两字符串最大公共字符串长度
*/
public static void main(String[] args) {
char[] s1 = "jiajiangayaoyao".toCharArray(); //测试数据
char[] s2 = "jiangyaoyao".toCharArray();
int c = new MaxConString().getCount(s1, s2);
System.out.println("两字符串的共同字符串长度为:"+c);
}
private int getSubCount(char[] s1,char[] s2, int i ,int j){//计算两字符串从s1的第i位置s2的第j位置的之后字符串长度
//如“abc”和“ab”则返回conut为2
int count=1;
while(++i<s1.length&&++j<s2.length&&s1[i]==s2[j]){
count++;
}
return count;
}
private int getCount(char[]s1,char[]s2){ //计算两字符串的共同字符串长度
int count = 0;
for(int i=0;i<s1.length;i++)
for(int j=0;j<s2.length;j++)
if(s1[i]==s2[j]){
if(this.getSubCount(s1, s2, i, j)>count)
count = this.getSubCount(s1, s2, i, j);
}
return count;
}
}6.1.将1-7个数字的全排列按照从小到大的顺序放在一个数组，例如第0个元素是1234567，第1个是1234576，第5039个是7654321.请问第1646个元素是多少？
答案：3265417.
7.6位数字且第一位不为0的美团券密码，在易个液晶数字设备上显示，倒过来看与原，ima一样的概率是多少（如129621）（液晶显示的数字1倒过来也算一样哦）。
8.求单链表的倒数第K个元素。
struct node
{
int key;
struct node* next;
};
typedef node* List;
实现该函数
int findLastKthElement(List list, int k)。
9.现有实数数组A和B，希望将A和B归并为一个有序数组C，且C中无重复的数，请写出算法并给出算法复杂度。
10.如果两个正整数a和b，a的所有真因子之和等于b，b的所有真因子之和等于a，则称a，b是amicable pair（说明：真因子包括1但不包括本身，比如14的真因子为1、2、7）。例如220和284就是amicable pair。
请写一段代码，打印出所有不超过1000万的amicable pair。