今天看到有人在問這個(gè)問題，寫了下代碼，標(biāo)準(zhǔn)庫分離了算法和數(shù)據(jù)結(jié)構(gòu)，按照這個(gè)框架寫程序確實(shí)比較方便，個(gè)人認(rèn)為熟讀和透徹理解標(biāo)準(zhǔn)庫源碼是每個(gè)想成為資深c++程序員的必修課，就框架結(jié)構(gòu)而論，stl很好的分離了算法和數(shù)據(jù)結(jié)構(gòu)，就算法而論，標(biāo)準(zhǔn)庫里有很多常見算法的經(jīng)典實(shí)現(xiàn)，所以有非常高的研究價(jià)值。
#include <iostream>
#include <stddef.h>
#include <stdlib.h>
#include <string>
#include <iterator>
#include <algorithm>
#include <vector>
using namespace std;
template <typename InputIterator1, typename InputIterator2, typename OutputIterator>
OutputIterator delete_intersection(InputIterator1 first1, InputIterator1 last1,
InputIterator2 first2, InputIterator2 last2, OutputIterator dest)
{
while (first1 != last1 && first2 != last2) {
if (*first1 > *first2) {
*dest = *first2;
++first2;
++dest;
} else if (*first1 < *first2) {
*dest = *first1;
++first1;
++dest;
} else {
++first1;
++first2;
}
}
for (;first2 != last2; ++first2) *dest = *first2;
return dest;
}
int main() {
int a[] = {1,1,2,2,5,6,9,9};
int b[] = {1,2,3,4,4,6,7,8,9,9,9,10};
vector<int> vc;
delete_intersection(a, a + sizeof(a)/sizeof(a[0]), b, b + sizeof(b)/sizeof(b[0]), back_inserter(vc));
std::copy(a, a + sizeof(a)/sizeof(a[0]), ostream_iterator<int>(cout, ","));
cout << endl;
std::copy(b, b + sizeof(b)/sizeof(b[0]), ostream_iterator<int>(cout, ","));
cout << endl;
std::copy(vc.begin(), vc.end(), ostream_iterator<int>(cout, ","));
cout << endl;
::system("PAUSE");
return EXIT_SUCCESS;
}posted on 2009-03-05 18:56
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