如何使用C++ std来计算整数集合的前导零和尾随零

Question

我们可以通过调用标准库来替换循环来计算整数集合中的前导零吗？

我正在学习std,但由于我需要知道前一个元素,所以无法想办法使用count或count_if之类的东西.

int collection[] = { 0,0,0,0,6,3,1,3,5,0,0 };
auto collectionSize = sizeof(collection) / sizeof(collection[0]);

auto countLeadingZeros = 0;
for (auto idx = 0; idx < collectionSize; idx++)
{
    if (collection[idx] == 0)
        countLeadingZeros++;
    else
        break;
}
// leading zeros: 4*0
cout << "leading zeros: " << countLeadingZeros << "*0" << endl;

我有一个类似的案例来统计同一个集合中的尾随零.

auto countTrailingZeros = 0;
for (auto idx = collectionSize - 1; idx >= 0; idx--)
{
    if (collection[idx] == 0)
        countTrailingZeros++;
    else
        break;
}
// trailing zeros: 2*0
cout << "trailing zeros: " << countTrailingZeros << "*0" << endl;

下面是一个完整的构建示例.

#include <iostream>

using namespace std;

int main()
{
    int collection[] = { 0,0,0,0,6,3,1,3,5,0,0 };
    auto collectionSize = sizeof(collection) / sizeof(collection[0]);

    auto countLeadingZeros = 0;
    for (auto idx = 0; idx < collectionSize; idx++)
    {
        if (collection[idx] == 0)
            countLeadingZeros++;
        else
            break;
    }
    cout << "leading zeros: " << countLeadingZeros << "*0" << endl;

    auto countTrailingZeros = 0;
    for (auto idx = collectionSize - 1; idx >= 0; idx--)
    {
        if (collection[idx] == 0)
            countTrailingZeros++;
        else
            break;
    }
    cout << "trailing zeros: " << countTrailingZeros << "*0" << endl;


    return 0;
}

Answer 1

一种方法是使用std::find_if.

auto countLeadingZeros = std::find_if(
    std::begin(collection), std::end(collection),
    [](int x) { return x != 0; }) - std::begin(collection);
auto countTrailingZeros = std::find_if(
    std::rbegin(collection), std::rend(collection),
    [](int x) { return x != 0; }) - std::rbegin(collection);