Question

所以我试图编写代码来对向量进行合并排序工作。然而，我现在遇到一个问题，即每当我输入矢量大小的大小不是2的幂时，最终的排序矢量将其大小向下舍入到最接近的2的幂。如果矢量大小设置为10，它将返回8，如果它是100，它将返回64.我不知道为什么会这样。这是代码：

#include <iostream>
#include <stdlib.h>     /* srand, rand */
#include <time.h>  /* time */
#include <vector>
#include <assert.h>
#include <iterator>
#include <algorithm> 
#include <functional> 
using namespace std;

vector<int> merge(vector<int> &left, vector<int> &right) {
    vector <int> result;
    while (left.size() > 0 && right.size()>0) {
        if (left[0] <= right[0]) {
            result.push_back(left[0]);
            left.erase(left.begin());
        }
        else {
            result.push_back(right[0]);
            right.erase(right.begin());
        }
    }
    if (left.size() > 0) {
        result.insert(result.end(), left.begin(), left.end());
    }
    if (right.size() > 0) {
        result.insert(result.end(), right.begin(), right.end());
    }
    return result;
}

vector <int> mergesort(vector < int > &vector_sort) {
    int size_of = vector_sort.size() / 2;
    vector <int > left(size_of), right(size_of), result(vector_sort.size());
    if (vector_sort.size() <= 1) {
        return vector_sort;
    }
    else {
        int middle = vector_sort.size() / 2;
        int counter = middle;
        int second_half = middle;
        for (int i = 0; i < middle; i++) {
            left[i] = vector_sort[i];
        }
        for (int i = 0; i < middle; i++) {
            right[i] = vector_sort[counter];
            counter++;
        }
        left = mergesort(left);
        right = mergesort(right);
        if (left.back() <= right[0]) {
            left.insert(left.end(), right.begin(), right.end());
            return left;
        }
        result = merge(left, right);
        return result;
    }
}
int main() {
    const int t=10;
    vector<int> vector_no(t);
    vector<int> empty(t);
    for (int i = 0; i < t; i++) {
        int hello = rand() % 100000 + 1;
        vector_no[i] = hello;
    }
    int hello2 = 0;
    for (int i = 0; i < vector_no.size(); i++) {
        hello2++; //I am using this for loop here just to count the size of the vector without getting a stack overflow error
    }
    cout << hello2  << "\n";
    vector_no = mergesort(vector_no);
    hello2 = 0;
    for (int i = 0; i < vector_no.size(); i++) {
        hello2++; //I am using this for loop here just to count the size of the vector without getting a stack overflow error
    }
    cout << hello2 << "\n";
}

Answer 1

当你构造右向量时，你应该添加一些等于middle的元素，当你应该添加vector_sort.size（） - 元素元素的中间数。因为除法向下舍入，这会导致您删除奇数长度的列表上的最后一个元素。并且因为您将向下排序合并为0或1个元素，所以它会使结果为2的幂。

合并排序仅返回大小为2的幂的向量

1 个答案: