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

#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:"<      for(int i=0;i<=L.length-1;i++)
         cout<      cout< }

 
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:"<     cin>>n;
    L.length = n;
    cout<<"please enter the element of your list:"<     for(int i=0;i<=L.length-1;i++){
       cin>>L.data[i];              
    }
    cout<<"the length of the list is:"<
}

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!"<     else
    {
        cout<<"now we insert "<         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!"<     else{
        obj = L.data[n-1];
        cout<<"now we delete the "<         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!"<     else{
        cout<<"now we modify the "<         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:"<     cin>>a;
    for(i=0;i<=L.length-1;i++)
    {
         if(L.data[i] == a){
             cout<<"the element "<              //找到就返回这个数的位置
             leap = 1;
             break;
             }
    }
    if(leap==0)
        cout<<"Sorry not found!"< }


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
OK！收工！