数据结构顺序表的“增删改查”功能实现

今天完成的是顺序表的简单功能实现

#include<iostream>

#define maxsize 100
using namespace std;
typedef struct{
        int data[maxsize];//data数组用来放数据 
        int length;//length是数组的长度 
}sqlist;
//建立一个结构体用来表示顺序表 


void printlist(sqlist &l)//打印顺序表 
{
     cout<<"now the list is:"<<endl;
     for(int i=0;i<=l.length-1;i++)
         cout<<l.data[i]<<" ";
     cout<<endl;
}

 
void initlist(sqlist &l)//初始化数组，即就是输入数组初始元素 
{
    int n;
    cout<<"please enter the number of your list:"<<endl;
    cin>>n;
    l.length = n;
    cout<<"please enter the element of your list:"<<endl;
    for(int i=0;i<=l.length-1;i++){
       cin>>l.data[i];              
    }
    cout<<"the length of the list is:"<<l.length<<endl;

}

void insertlist(sqlist &l,int n,int p)//插入数据操作 
{
    int i;
    if(n<0 || n>l.length)
        cout<<"error!please reput!"<<endl;
    else
    {
        cout<<"now we insert "<<p<<" to the "<<n<<" th position in the"<<" list"<<endl;
        for(i=l.length-1;i>=n-1;--i)
            l.data[i+1] = l.data[i];
            //先把插入点之后的所有元素依次向后移动一位 
        l.data[n-1] = p;
        //然后把目标元素插入到目标点当中 
        l.length += 1;//数组长度加一 
    }
}

void deleteelem(sqlist &l,int n)//删除数组中的元素 
{
    int i,obj;
    if(n<0 || n>l.length)
        cout<<"error!"<<endl;
    else{
        obj = l.data[n-1];
        cout<<"now we delete the "<<n<<" th position element "<<obj<<" in the list"<<endl;
        for(i=n;i<=l.length;i++)
            l.data[i-1] = l.data[i];//将删除点之后的元素都向前移一位     
        l.length -= 1;//数组长度减去一 
    }          
}

void modifylist(sqlist &l,int n,int p)//修改数组元素 
{
    if(n<0 || n>l.length-1)
        cout<<"error!"<<endl;
    else{
        cout<<"now we modify the "<<n<<" th element into the element "<<p<<endl;
        l.data[n-1] = p;//很简单直接赋值就好 
    }     
}

int findele(sqlist l)//查找数组元素 
{
    int i,a,leap=0;
    cout<<"please enter the element you want to find:"<<endl;
    cin>>a;
    for(i=0;i<=l.length-1;i++)
    {
         if(l.data[i] == a){
             cout<<"the element "<<a<<" is in "<<i+1<<"th position in the list"<<endl;          
             //找到就返回这个数的位置 
             leap = 1;
             break;
             }
    }
    if(leap==0)
        cout<<"sorry not found!"<<endl;
}


int main(){
    //主函数在此 
    sqlist p;
    initlist(p);
    printlist(p);
    insertlist(p,2,4);
    printlist(p);
    deleteelem(p,3);
    printlist(p);
    modifylist(p,4,18);
    printlist(p);
    findele(p);    
    system("pause");
    return 0;
    //system("pause")；一定要在return之前 
}

输出展示：

一样，python再来实现一遍：

#-*-coding:utf-8-*-
a = map(int,raw_input("enter the list:\n").split())
print 'the list is:\n'
print a
print 'the length of the list is:',len(a)
def add(list):
    print 'enter the position and number you want to add: '
    m,n = map(int,(raw_input().split()))
    b = list[:m-1]
    c = list[m-1:]
    b.append(n)
    d = b + c
    print "the list is ",d,' now'
    return d
a = add(a)
print 'enter the position you want to remove in the list'
k= int(raw_input())
a.remove(a[k-1])
print "the list is ",a,' now'
p,q =map(int,raw_input("enter the position and number you want to modify:\n").split())
a[p-1] = q
print "the list is ",a,' now'
print 'enter the number you want to find in the list'
n = int(raw_input())
if n in a:
    for i in range(len(a)):
        if a[i] == n:
            print "%d is the %dth element in the list"% (n,i+1),a
else:
    print 'not found!\n'

输出展示：

e:\python27\python.exe d:/python/python算法/b.py
enter the list:
1 2 3 4
the list is:

[1, 2, 3, 4]
the length of the list is: 4
enter the position and number you want to add: 
1 2
the list is  [2, 1, 2, 3, 4]  now
enter the position you want to remove in the list
2
the list is  [2, 2, 3, 4]  now
enter the position and number you want to modify:
1 5
the list is  [5, 2, 3, 4]  now
enter the number you want to find in the list
4
4 is the 4th element in the list [5, 2, 3, 4]

process finished with exit code 0

收工！