noo*_*guy 4 c++ arrays algorithm iterator min
我有两个数组:
int playerSums[9] = { };
string playerNames[9] = { };
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我试图获取数组中的最小值playerSums以及该值的数组索引。
到目前为止,这是我尝试过的:
if (playerNames[index] == "End" || playerNames[index] == "end") {
int lowestValue = playerSums[0];
for (i = 1; i < sizeof(playerSums) / sizeof(playerSums[0]); i++) {
if (playerSums[i] < lowestValue || lowestValue != 0)
lowestValue = playerSums[i];
}
cout << index[playerNames] << " had the lowest values and got the sum ";
cout << lowestValue << endl;
}
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playerSums例如,如果只有 3 个玩家在玩,即仅填充了数组的 3 个元素(其余元素等于 0),如何查找并显示数组中的最小值?
我需要索引来显示获得最小值的玩家的姓名。
您可以使用std::min_element标头中声明的标准算法<algorithm>来查找具有最小和的元素。例如
#include <algorithm>
int *min = std::min_element( playerSums, playerSums + 3 );
std::cout << playerNames[min - playerSums]
<< " had the lowest values and got the sum " << *min
<< std::endl;
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可以使用标准函数编写相同的内容std::begin,std::end并std::distance在标头中声明<iterator>
#include <algorithm>
#include <iterator>
int *min = std::min_element( std::begin( playerSums ), std::end( playerSums ) );
std::cout << playerNames[ std::distance( playerSums, min )]
<< " had the lowest values and got the sum " << *min
<< std::endl;
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您可以编写类似于该算法的自己的函数,而不是使用该算法。例如
size_t min_sum( const int playerSums[], size_t n )
{
size_t min = 0;
for ( size_t i = 1; i < n; i++ )
{
if ( playerSums[min] < playerSums[i] ) min = i;
}
return min;
}
size_t min = min_sum( playerSums, sizeof( playerSums ) / sizeof( *playerSums ) );
std::cout << playerNames[min]
<< " had the lowest values and got the sum " << playerSums[min]
<< std::endl;
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如果您需要跳过数组中等于零的元素,那么该函数将如下所示
size_t min_sum( const int playerSums[], size_t n )
{
size_t min = 0;
while ( min < n && playerSums[i] == 0 ) ++min;
for ( size_t i = min; i < n; i++ )
{
if ( playerSums[min] < playerSums[i] ) min = i;
}
return min;
}
size_t min = min_sum( playerSums, sizeof( playerSums ) / sizeof( *playerSums ) );
if ( min != sizeof( playerSums ) / sizeof( *playerSums ) )
{
std::cout << playerNames[min]
<< " had the lowest values and got the sum " << playerSums[min]
<< std::endl;
}
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像往常一样,最简单的解决方案是使用标准库,例如
auto it = std::min_element(std::begin(playerSums), std::end(playerSums));
std::size_t index = std::distance(std::begin(playerSums), it);
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现在您可以通过取消引用迭代器来获取最小值it:
int lowestValue = *it;
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如果您只想迭代数组中的前 3 个元素,那么您可以这样做:
auto first = std::begin(playerSums);
auto it = std::min_element(first, std::next(first, 3));
std::size_t index = std::distance(first, it);
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注意:更喜欢std::next而不是普通的指针算术(例如playerSums + 3),因为它更通用(适用于所有迭代器类型)。