查找數組最小「對」

我有「陣列偶」的數組：查找數組最小「對」

[[time, distance], [time, distance], [time, distance], [time, distance], ...]

我想找到「對」與最小時間的索引，如果有多個「配對'的最小時間相同，以最小距離取對。

例如如果我有：

[[5, 5], [1,7], [2,6], [1,6]]

我想返回3.（即第4個元素的索引）。

我該怎麼做？

來源

2014-12-13 user3711600

Array#sort確實如你所願。

myarray = [[5, 5], [1,7], [2,6], [1,6]] 
myarray.sort 
#=> [[1, 6], [1, 7], [2, 6], [5, 5]]

所以，我們只需要在原始數組中匹配排序數組中第一個元素的元素。

ndx = myarray.index myarray.sort[0] 
#=> 3

編輯：我最初使用太空飛船運營商，但意識到這是沒有必要的。

來源

2014-12-13 04:06:53 Fred

我不熟悉Ruby on Rails，但我寫了一點PHP代碼來實現你的目標。我希望你能從中提取出想法。

<?php 

$timesAndDistances = [['t' => t1, 'd' => d1], ['t' => t2, 'd' => d2], ..., ['t' => tn, 'd' => dn]]; 

$lowestTime_withDistance = [['t' => null, 'd' => null, 'i' => null]]; 

foreach($timesAndDistances as $index => $timeAndDistance) 
{ 
    $time = $timeAndDistance['t']; 
    $distance = $timeAndDistance['d']; 

    if($lowestTime_withDistance[0]['t'] == null) 
    { 
     $lowestTime_withDistance[0] = ['t' => $time, 'd' => $distance, 'i' => $index]; 
    } 
    else 
    { 
     if($lowestTime_withDistance[0]['t'] > $time){ 
      $lowestTime_withDistance[0] = ['t' => $time, 'd' => $distance, 'i' => $index]; 
     } 
     elseif ($lowestTime_withDistance[0]['t'] == $time) 
     { 
      $lowestTime_withDistance[] = ['t' => $time, 'd' => $distance, 'i' => $index]; 
     } 
    } 
} 

if(count($lowestTime_withDistance) == 1) 
{ 
    echo 'Index of element with minimum time is ' . $lowestTime_withDistance[0]['i']; 
} 
elseif(count($lowestTime_withDistance) > 1){ 
    $lowestDistance = [['d' => null, 'i' =>null]]; 

    foreach($lowestTime_withDistance as $timeWithDistance) 
    { 
     if($lowestDistance[0]['d'] == null) 
     { 
      $lowestDistance[0] = ['d' => $timeWithDistance['d'], 'i' => $timeWithDistance['i']]; 
     } 
     else 
     { 
      if($lowestDistance[0]['d'] > $timeWithDistance['d']) 
      { 
       $lowestDistance[0] = ['d' => $timeWithDistance['d'], 'i' => $timeWithDistance['i']]; 
      } 
      elseif ($lowestDistance[0]['d'] == $timeWithDistance['d']) 
      { 
       $lowestDistance[] = ['d' => $timeWithDistance['d'], 'i' => $timeWithDistance['i']]; 
      } 
     } 
    } 

    if(count($lowestDistance) == 1) 
    { 
     echo 'Index of element with minimum time and distance is ' . $lowestDistance[0]['i']; 
    } 
    elseif (count($lowestDistance) > 1) 
    { 
     foreach($lowestDistance as $aLowestDistance) 
     { 
      echo 'Index of an element with minimum time and distance is ' . $aLowestDistance['i']; 
     } 
    } 
}

來源

2014-12-13 04:09:24 budhajeewa

寫入快速代碼來實現這一點：

說陣列一個具有下列元素

[[5,5]，[1,7]，[2,6 ]，[1,6]]
創建陣列的b中的副本：

B = A

B = [[5,5]，[1,7]，[2,6]，[1,6]]
排序b

b = b.sort

b = [[1,6]，[1,7]，[2,6]，[5,5]]
遍歷嵌套循環

最小= B [0]

索引= 0

a.each做| iter_orig |除非a.nil？

if iter_orig.eql？最小

break;

end

index ++;

末

末

看跌期權「指數最小的時間/距離 - 」 +指數

來源

2014-12-13 04:17:35

查找數組最小「對」

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