合并排序算法辅助

时间:2013-09-14 20:58:22

标签: c++ algorithm mergesort

我正在尝试实现合并排序算法。我开始使用算法手册中提供的伪代码。伪代码表示数组中的第一个位置为1而不是0.我在尝试实现代码时非常困难。

这就是我所拥有的。我已经尝试通过在每一步打印出结果来逐步完成递归,但此时它非常复杂。

#include <iostream>
#include <deque>
using size_type = std::deque<int>::size_type;
void print(std::deque<int> &v)
{
    for(const auto &ref:v)
        std::cout << ref << " ";
    std::cout << std::endl;
}
void merge(std::deque<int> &vec, size_type p, size_type q, size_type r)
{
    int n_1 = q - p;
    int n_2 = r - q;
    std::deque<int> left, right;
    for(auto i = 0; i != n_1; i++)
        left.push_back(vec[p + i]);
    for(auto j = 0; j != n_2; j++)
        right.push_back(vec[q + j]);
    int i = 0, j = 0;
    std::cout << "left = ";
    print(left);
    std::cout << "right = ";
    print(right);
    for(auto k = p; k != r; k++) {
        if((i != n_1 && j != n_2) && left[i] <= right[j]) {
            vec[k] = left[i];
            i++;
        }
        else if(j != n_2){
            vec[k] = right[j];
            j++;
        }
    }
}
void merge_sort(std::deque<int> &A, size_type p, size_type r)
{
    int q;
    if(p < r - 1) {
        q = (p + r)/2;
        merge_sort(A, p, q);
        merge_sort(A, q + 1, r);
        merge(A, p, q, r);
    }
}
int main()
{
    std::deque<int> small_vec = {1, 6, 2, 10, 5, 2, 12, 6};
    std::deque<int> samp_vec = {2, 9, 482, 72, 42, 3, 4, 9, 8, 73, 8, 0, 98, 72, 473, 72, 3, 4, 9, 7, 6, 5, 6953, 583};
    print(small_vec);
    merge_sort(small_vec, 0, small_vec.size());
    print(small_vec);
    return 0;
}

运行程序时,我得到以下输出:

left = 1 
right = 6 
left = 1 6 
right = 2 10 
left = 2 
right = 12 6 
left = 1 2 6 10 
right = 5 2 12 6 
1 2 5 2 6 10 12 6 

2 个答案:

答案 0 :(得分:1)

错误在于:(i != n_1 && j != n_2) && left[i] <= right[j])i != n_1评估为false vec[k] = right[j];时执行 - 正确。

但如果i != n_1评估为truej != n_2评估为假,j = n_2您的程序会再次尝试vec[k] = right[j];,即访问您的界限双端队列。

重写你的for循环,如下所示: if (i<n_1 && (j>=n_2 || left[i] <= right[j]) 此循环仅由于C ++的条件短路而起作用,即当j>=n_2求值为true时,left[i] <= right[j]永远不会再次检查,并且您不会访问deque over bounds。

仅当left[i] <= right[j]为真且i<n_1为假时才会检查

j>=n_2,否则执行第二个分支。

答案 1 :(得分:1)

花了很多时间在另一篇帖子上获得一些有价值的帮助后,才能让算法正常运行。

正确代码:

#include <iostream>
#include <deque>
using size_type = std::deque<int>::size_type;
void print(std::deque<int> &v)
{
    for(const auto &ref:v)
        std::cout << ref << " ";
    std::cout << std::endl;
}
void print(int arr[], int size)
{
    for(int i = 0; i != size; i++)
        std::cout << arr[i] << " ";
    std::cout << std::endl;
}
void merge(std::deque<int> &vec, size_type p, size_type q, size_type r)
{
    int n_1 = q - p + 1;
    int n_2 = r - q;
    std::deque<int> left, right;
    int i = 0, j = 0;
    while(i < n_1)
        left.push_back(vec[p + i++]);
    while(j < n_2)
        right.push_back(vec[j++ + q + 1]);
    i = 0; j = 0;
    //std::cout << "left = ";
    //print(left);
    //std::cout << "right = ";
    //print(right);
    for(auto k = p; k <= r; k++) {
        if((i < n_1 && left[i] <= right[j]) || j >= n_2) {
            vec[k] = left[i++];
        }
        else if(j < n_2){
            vec[k] = right[j++];
        }
    }
}
void merge_sort(std::deque<int> &A, size_type p, size_type r)
{
    int q;
    if(p < r) {
        q = (r + p) / 2;
        std::cout << "q = " << q << std::endl;
        //std::cout << "p = " << p << std::endl;
        merge_sort(A, p, q);
        merge_sort(A, q + 1, r);
        merge(A, p, q, r);
    }
}
int main()
{
    std::deque<int> small_vec = {10, 3, 6, 4, 1, 5, 3, 9, 7, 2, 8};
    std::deque<int> samp_vec = {2, 9, 482, 42, 3, 4, 9, 8, 73, 8, 0, 98, 72, 473, 72, 3, 4, 9, 7, 6, 5, 6953, 583};
    print(samp_vec);
    merge_sort(samp_vec, 0, samp_vec.size() - 1);
    print(samp_vec);
    return 0;
}