c++ - 在数组中查找重复项

问题描述

我对这种语法一无所知int *count = new int[sizeof(int)* (size - 2)]

将创建什么样的数组。

我以为他们正在尝试创建类似地图的结构。但它是如何工作的？

#include <bits/stdc++.h>
using namespace std;
 
void printRepeating(int arr[], int size)
{
    int *count = new int[sizeof(int)*(size - 2)];
    int i;
         
    cout << " Repeating elements are ";
    for(i = 0; i < size; i++)
    {
        if(count[arr[i]] == 1)
            cout << arr[i] << " ";
        else
            count[arr[i]]++;
    }
}
 
// Driver code
int main()
{
    int arr[] = {4, 2, 4, 5, 2, 3, 1};
    int arr_size = sizeof(arr)/sizeof(arr[0]);
    printRepeating(arr, arr_size);
    return 0;
}
 
// This is code is contributed by rathbhupendra

标签： c++

在数组中查找重复项当然是一个已解决的问题。恕我直言，一个非常简单的解决方案是：

使用 std::sort() 对数组进行排序
使用循环来检查一个元素是否等于它的后继元素，即。for(int i = 1; i < num_elements; i++){ if(arr[i-1]==arr[i]){...duplicate!...}}

这需要 O(n) 内存和 O(n*log(n)) 时间，所以没问题。您也可以使用哈希图，但这几乎是一样的。

无论如何，对于您的问题：

int *count = new int[sizeof(int)* (size - 2)];

这是不正确的。我假设它曾经是这个 C 代码：

int num_elements = size-2; // we want size-2 elements (not sure why)
int total_bytes = sizeof(int) * num_elements; 
int *count = calloc(total_bytes); // reserve space, and set to 0

哪一个可以翻译成这个 C++ 代码：

int num_elements = size-2; 
int * count = new int[num_elements]{0}; // alloc and set to zero

所以移植的人误解了 C++ 的基本原理。

让我们进一步挖掘。

为此，问题表述似乎是：

您将获得一个包含 n+2 个元素的数组。数组的所有元素都在 1 到 n 的范围内。并且所有元素都出现一次，除了两个出现两次的数字。找到两个重复的数字

我进行了微小的更改以使解决方案不那么疯狂，并且添加了注释。

// #include <bits/stdc++.h> is a bad choice. 
// This includes _EVERYTHING_ in C++, 
// but it only works in GCC afaik. For this particular 
// case we just need cout, so: 
#include <iostream>

// using the std namespace like this spills function calls like crazy in the global namespace. 
// it is both, conventient and "not too bad" (imho) in cpp files, 
// but never ever do this in .h files where it affects multiple cpp files. 
using namespace std;
 
void printRepeating(int arr[], int size)
{
    // create a new array of size-2 and set to zero
    int *count = new int[(size - 2)]{0};
    int i;
         
    cout << " Repeating elements are ";

    // loop all elements
    for(i = 0; i < size; i++)
    {
        int value = arr[i]; 

        // increase the count for `value`
        // note that if value<=0 or value>size-2, 
        // then the program will crash! 
        // the problem description is contrived, but it does 
        // state that all values need to be >0 and <=size-2, 
        // so this next array access is fine! 
        count[value-1]++;
        // we still have to subtract 1, because C arrays start 
        // at index 0, but our problem description says we start at 1. 
        // we could also create an array of size-1 and 
        // "ignore" the first position in count[0] (it would never get used!)


        // if the element appears the second time... 
        if(count[value-1] == 2){
            // then print it
            cout << value << " ";
            // btw: if you check with count[value-1]==2, then only
            // the first duplicate is printed. 
            // you could compare with count[value-1]>=2 then all 
            // repeating elements are printed repeatetly. 
        }
    }

    // free up the memory. we allocated with `new[]`
    // so we also have to use `delete[]`
    delete [] count; 
}
 
// Driver code
int main()
{
    int arr[] = {4, 2, 4, 5, 2, 3, 1};
    // next line is a standard trick. 
    // a single int consumes 4bytes (typically), 
    // the array will have size 7*4=28 bytes, so sizeof(arr)=28
    // the first element is an int, so sizeof(arr[0]) = 4
    // so sizeof(arr)/sizeof(arr[0]) = 7
    // c and c++ don't have arr.length like java,etc., that's 
    // why under certain circumstances this "trick" is used. 
    int arr_size = sizeof(arr)/sizeof(arr[0]);
    // call the function
    printRepeating(arr, arr_size);
    return 0;
}
 
// This is code is contributed by rathbhupendra. 
// Maybe fixed and annotated by hansi:)

我希望这对您有所帮助并回答一些问题。祝你的 C++ 冒险好运！

c++ - 在数组中查找重复项 - 语法说明

问题描述

解决方案

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