删掉重复的数据

 

一）过滤重复的数据。即把原来重复的数据过滤，只保留一份。

例如：int a[]={1,3,4,3,2,5,2};
输出：1 3 4 2 5

①实现方法一(采用STL)：

#include <iostream>
#include <algorithm>
#include <set>

using namespace std;

void Dedup(FILE* infile,FILE* outfile)
{
    int iTemp;
    unsigned int nCount = 0;

    typedef set<int> IntSet;
    IntSet iArray;

    while(NULL == feof(infile))
    {
        fscanf(infile,"%d",&iTemp);
        iArray.insert(iTemp);

        nCount ++;
    }

    printf("nCount = %d\n",nCount);//total numbers

    for(IntSet::iterator iter = iArray.begin(); iter != iArray.end(); ++ iter)
    {
        fprintf(outfile,"%d ",*iter);
    }

}

int main()
{
    FILE *fpInput,*fpOutput;

    if(NULL == (fpInput = fopen("D:\\input.txt","r")))
    {
        printf("File read error!\n");
        exit(1);
    }

    if(NULL == (fpOutput = fopen("D:\\output.txt","w")))
    {
        printf("File write error!\n");
        exit(1);
    }

    Dedup(fpInput,fpOutput);
    printf("\n");

    fclose(fpInput);
    fclose(fpOutput);

    return 0;
}

②实现方法二(采用STL)：

#include <iostream>
#include <algorithm>
#include <vector>

using namespace std;

void Dedup(FILE* infile,FILE* outfile)
{
    int iTemp;
    unsigned int nCount = 0;

    vector<int> iArray,iRes;

    while(NULL == feof(infile))
    {
        fscanf(infile,"%d",&iTemp);
        iArray.push_back(iTemp);

        nCount ++;
    }

    printf("nCount = %d\n",nCount);//total numbers

    sort(iArray.begin(), iArray.end());//sort

    unique_copy(iArray.begin(), iArray.end(),back_inserter(iRes));//去掉重复的

    for(vector <int>::const_iterator iter = iRes.begin(); iter != iRes.end(); ++ iter)
    {
        fprintf(outfile,"%d ",*iter);
    }
}
int main()
{
    FILE *fpInput,*fpOutput;

    if(NULL == (fpInput = fopen("D:\\input.txt","r")))
    {
        printf("File read error!\n");
        exit(1);
    }

    if(NULL == (fpOutput = fopen("D:\\output.txt","w")))
    {
        printf("File write error!\n");
        exit(1);
    }

    Dedup(fpInput,fpOutput);
    printf("\n");

    fclose(fpInput);
    fclose(fpOutput);

    return 0;
}

③实现方法三(采用STL)：

#include   <iostream>
#include   <string>
#include   <vector>
#include   <set>
#include   <algorithm>

using namespace std;

template <typename T>
void print_element(T value)
{
    cout<<value<<" ";
}

template <typename T>
unsigned int MakeUniqueArray1(T* array, unsigned int length)
{
    set<T> s(array, array + length);
    //copy(s.begin(), s.end(), array);

    printf("%d Number1.\n",s.size());
    return s.size();
}

template <typename T>
unsigned int MakeUniqueArray2(T* array, unsigned int length)
{
    vector<T> v(array, array + length);
    sort(v.begin(), v.end());

    T* end = unique_copy(v.begin(), v.end(), array);

    printf("%d Number2.\n",end - array);

    return end - array;
}

int main()
{
    int a[] = {0,1,2,3,4,5,6,7,8,9,0,9,8,7,6,5,4,3,2,1};

    unsigned int len = sizeof(a) / sizeof(int);

    len = MakeUniqueArray1(a, len);
    len = MakeUniqueArray2(a, len);

    void (*pfi)(int) = print_element;
    for_each(a, a + len, print_element<int>);

    printf("\n");
    return 0;
}

④C++ 普通实现方法

#include <stdio.h>

int* XRemove(int *Src, int Size,int *Dst, int &nSize)
{
    bool* Index = new bool[Size];

    for(int i = 0; i < Size; ++i)
        Index[i] = true;

    int* Crt = Dst;

    nSize = 0;
    for(int x = 0; x < Size; ++x)
    {
        if(0 == Index[x])
            continue;

        for(int y = x + 1; y < Size; ++y)
            if(Src[x] == Src[y])
                Index[y] = false;

        if(Index[x]){
            nSize ++;
            *Crt++ = Src[x];
        }
    }

    delete [] Index;
    return Dst;
}

void main()
{
    int Src[]={1,3,4,3,2,5,2};

    int Size = sizeof(Src) / sizeof(int);
    int *Dst =  new int [Size];

    int nSize;
    XRemove(Src, Size, Dst, nSize);

    for(int i = 0;i < nSize;i ++)
        printf("%d ",Dst[i]);

    printf("\n");
}

二）去掉重复出现的数字

例如：int a[]={1,3,4,3,2,5,2};

输出：1 4 5

C++ 实现方法：

#include <stdio.h>

int* XRemove(int Dst[], int Src[], int Size,int &nSize)
{
    bool* Index = new bool[Size];

    for(int i = 0; i < Size; ++i)
        Index[i] = true;

    int* Crt = Dst;
    bool Sig;

    nSize = 0;
    for(int x = 0; Sig = (x < Size); ++x)
    {
        if(Index[x] == 0)
            continue;

        for(int y = x + 1; y < Size; ++y)
            if(Src[x] == Src[y])
                Sig = Index[y] = false;

        if(Sig){
            nSize ++;
            *Crt++ = Src[x];
        }
    }

    delete [] Index;
    return Dst;
}

void main()
{
    int Src[]={1,3,4,3,2,5,2};

    int Size = sizeof(Src) / sizeof(int);

    int *Dst =  new int [Size];

    int nSize;

    XRemove(Dst,Src, Size,nSize);

    for(int i = 0;i < nSize;i ++)
        printf("%d ",Dst[i]);

    printf("\n");
}

小结：使用vector容器可以很快解决。看来还是要多掌握一些知识和技术，拓宽自己的研究。有些问题也许用另一种方法，其实很简单，执行效率也很高。那么，我们为何不尝试一下呢？从中想想，要注意的问题？

 

来源：http://blog.csdn.net/tianmohust/article/details/7027713