使用C++容器vector實現的兩路歸併排序

這篇文章的目的是爲了讓後來者，也讓我自己，記住容器vector的一些操作。

RT.順便實現了計算number of inversions（其實這纔是重點）。

 

/*Definition:Let A[1..n] be an array of 'n' distinct numbers. If i < j and A[i] > A[j], then the pair (i,j) is called an inversion of A.
*/
#include <iostream>
#include <stdlib.h>
#include <vector>
#include <fstream>
#include <ctime>
using namespace std;
long long inversionNum = 0;
vector<int> mergeSort(vector<int> &i);
vector<int> merge(vector<int> &left, vector<int> &right);
int main(){
time_t begin,end;     begin=clock();
    //read file,initial a integer vector
    vector<int> ivec(100000,0);
    vector<string> svec(100000);
    ifstream fin("beforeSort");
    for (vector<string>::iterator sx = svec.begin(); sx != svec.end(); ++sx)
        getline(fin, *sx);
    fin.close();
    //change it into integer
    for (int count = 0; count < 100000; ++count)
        ivec[count] = atoi(svec[count].c_str());
    //merge sort

    vector<int> result = mergeSort(ivec);
    ofstream fout("afterSort");
    for(vector<int>::iterator rx = result.begin(); rx != result.end(); ++rx)
        fout << *rx << endl;
    fout.close();

    cout << "The number of inversions is:" << inversionNum << endl;

end=clock();
cout<<"runtime: "<<double(end-begin)/CLOCKS_PER_SEC<<endl;
    return 0;
}

vector<int> mergeSort(vector<int>& i) {
    if ( i.size() == 1)
        return vector<int >(1, i[0]);
    if ( i.size() == 0){
        return vector<int >();
    }

    vector<int>::iterator mid = i.begin() + i.size() / 2;
    vector<int> left(i.begin(), mid);
    vector<int> right(mid, i.end());

    left = mergeSort( left );
    right = mergeSort( right );

    return merge( left, right);
}

vector<int> merge(vector<int> &left, vector<int> &right) {
    vector<int> result;
    vector<int>::size_type li = 0, ri = 0;
    while ( li < left.size() && ri < right.size()) {
        if ( left[li] > right[ri] ) {   //inversion
            result.push_back(right[ri++]);
            inversionNum += (left.size() - li);
        }
        else {
            result.push_back(left[li++]);
        }
    }
    while ( li < left.size() ) {
        result.push_back(left[li++]);
    }
    while ( ri < right.size() ) {
        result.push_back(right[ri++]);
    }
    return result;
}