完美世界笔试题之破解密电前需要获取完整的数字电报,将电报里的数字反序同时还需要去除掉多余的空格;//例如:” 1 5721 23 “–>”23 5721 1”
我的思路:
- 1,去掉开头和 末尾的空格
- 2,处理字符串中间的多余空格,(用后面的字符填充)
- 3,旋转处理:先逆置整个字符串,再把每个字串逆置
代码如下:
#include<iostream>
#include<string>
#include<string.h>
#include<stdio.h>
using namespace std;
void Reverse(string& str,int start,int end)
{
while (start < end && start != end + 1)
{
swap(str[start++], str[end--]);
}
}
void fun(string& str)
{
//处理开头和末尾的空格
int start = 0;
int end = str.size() - 1;
while (start < str.size())
{
if (str[start] != ' ')
break;
++start;
}
while (end >= 0)
{
if (str[end] != ' ')
break;
--end;
}
str = str.substr(start, end + 1);
//再处理多余的空格,只保留一个
start = 0;
end = 0;
while (end < str.size())
{
if (str[end] != ' ')
{
str[start++] = str[end];
}
if (str[end] == ' ' && str[end - 1] != ' ')
{
str[start++] = str[end];
}
++end;
}
str = str.substr(0, start);
//此时str空格的问题全部解决,剩下顺序的问题,采用三步旋转解决(旋转整个字符串,再旋转每个字串)
start = 0;
end = 0;
Reverse(str, start, str.size()-1);
for (int i = 0; i < str.size(); ++i)
{
if (str[i] == ' ')
{
Reverse(str, start, i - 1);
start = i + 1;
}
}
}
int main()
{
string str;
getline(cin, str);
fun(str);
cout << str.c_str() << endl;
system("pause");
return 0;
}
双向链表实现队列:
struct listNode
{
int data;
listNode* next;
listNode* prev;
listNode(int x)
{
data = x;
next = NULL;
prev = NULL;
}
};
class myqueue
{
public:
myqueue()
:head(NULL)
, tail(NULL)
{}
void Push(int x)
{
listNode* node = new listNode(x);
if (head == NULL)//第一次插入
{
head = node;
tail = node;
}
else
{
node->prev = tail;
tail->next = node;
tail = node;
}
}
void Pop()
{
if (head == NULL)
{
cout << "error:the queue is NULL" << endl;
return;
}
if (head->next == NULL)
{
delete(head);
head = NULL;
tail = NULL;
}
else
{
head = head->next;
delete(head->prev);
head->prev = NULL;
}
}
int Top()
{
if (head == NULL)
{
cout << "error:the queue is NULL" << endl;
return;
}
else
return head->data;
}
private:
listNode* head;
listNode* tail;
};
int main()
{
myqueue q;
q.Push(1);
q.Push(2);
q.Push(3);
q.Pop();
q.Pop();
q.Pop();
system("pause");
return 0;
}