查找陣列

Question

最小值我有兩個數組：

int playerSums[9] = { }; 
string playerNames[9] = { }; 

我試圖讓陣列playerSums在最小值，也該值的數組索引。

這裏是我試過到目前爲止：

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; 
} 

如何找到和陣列playerSums中顯示的最小值例如如果只有3名球員被打，即僅3元數組被填充（並且其餘的元素等於零）？

我需要索引來顯示獲得最小值的玩家的名字。

Answer 1

您可以使用標準算法std::min_element在頭<algorithm>宣稱找到元素WITN最小總和。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; 

同樣可以使用標準功能std::begin，std::end和std::distance使用的算法，您可以編寫類似的算法自己功能的頭部聲明<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; 

而是被寫入。例如

size_t min_sum(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; 

如果您需要跳過等於零，則數組元素的功能將類似於

size_t min_sum(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; 
} 

Answer 2

你知道你分配給lowestValue當您更改該變量的值的元素的索引，所以只保存指數的變量（比如，index），這樣，當你完成index擁有的最後一個值的索引分配。

Answer 3

首先調整你的循環條件。我不確定你以前是否定義過我，所以你可能忘記了。第二個停止條件是sizeof（palyerSums）就足夠了。你也只需要存儲數組中最低的playerSums的索引。 if條件也有太多的東西。如果最低值不爲零，您將始終更改該值，除非最低值恰好爲零，否則這看起來不正確。

int lowestValue = playerSums[0]; 
int resultIndex = 0; 
for(int i = 1; i < sizeof(playerSums)/sizeof(playerSums[0]); i++) { 
    if(playerSums[i] < lowestValue) { 
    lowestValue = playerSums[i]; 
    resultIndex = i; 
    } 
} 
cout << playerNames[resultIndex] << "blabla" << lowestValue; // instead of lowestValue you could also do playerSums[resultIndex] ofcourse. 

讓我知道是否可行

Answer 4

您存儲lowestValue值最低的同樣的方式，存儲索引在一個變量中，比方說，lowestValueIndex。此外，刪除外，如果和移動它內部的for循環：

if(playerNames[i] == "End" || playerNames[i] == "end") 
    break; 

這樣，您將確保只有誰在玩的玩家將被處理。此外，您不需要檢查最低值是否爲零。因此，代碼會看起來像：

int lowestValue = playerSums[0]; 
int lowestValueIndex = 0; 
for (int i = 1; i < sizeof(playerSums)/sizeof(playerSums[0]); ++i) 
{ 
    if(playerNames[i] == "End" || playerNames[i] == "end") 
     break; 
    if (playerSums[i] < lowestValue) 
    { 
      lowestValue = playerSums[i]; 
      lowestValueIndex = i; 
    } 
} 
cout << index[playerNames] << " had the lowest values and got the sum " 
    << lowestValue << endl; 

正如一個音符，使用可以長到簡化這個（如vector）的標準數組：

std::vector<std::string> playerNames; 
std::vector<int> playerSums; 

for (int i = 1; i < playerSums.size(); ++i) 
{ 
    if (playerSums[i] < lowestValue) 
    { 
      lowestValue = playerSums[i]; 
      lowestValueIndex = i; 
    } 
} 
cout << index[playerNames] << " had the lowest values and got the sum " 
    << lowestValue << endl; 

Answer 5

像往常一樣，最簡單的解決方法是使用標準庫，例如

int lowestValue = *it; 

如果只想遍歷數組中的第3個元素，那麼你可以做這樣的事情：

auto it = std::min_element(std::begin(playerSums), std::end(playerSums)); 
std::size_t index = std::distance(std::begin(playerSums), it); 

現在，您可以通過提領該迭代it得到最小值而不是：

auto first = std::begin(playerSums); 
auto it = std::min_element(first, std::next(first, 3)); 
std::size_t index = std::distance(first, it); 

注：喜歡std::next而不是簡單的指針算術（例如playerSums + 3），因爲它更通用（適用於所有迭代器類型）。