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幫幫這個簡單的演算法

WBOY
WBOY原創
2016-09-19 09:16:25998瀏覽

寫的不好,請大家優化
一個用戶ID的陣列

<code>$uids = [1,2,3,5,6,8,9,11,13,25,65];</code>

這個陣列的每一個鍵值代表一個UID
金額數組

<code>$amounts = [
    1=>12000,
    2=>500,
    3=>11000,
    5=>1000,
    6=>11000,
    8=>12000,
    9=>12000,
    11=>11000,
    13=>12000,
    25=>22000,
    65=>123123
];</code>

此數組的鍵名和$uid數組的鍵值對應。
循環$uids這個數組,取出對應的金額$amount,
如果這個金額大於等於12000($boundary),就在總金額($totals)加上這個$amount
如果這個金額小於12000,就再向下循環,循環到這幾個$amount相加大於等於12000,然後在總金額($totals)加上這幾個$amount的和
最多取三層。
最後得出$totals的值。
我現在的程式碼:

<code><?php
$boundary = 12000;
$uids= [1,2,3,5,6,8,9,11,13,25,65];
$amounts = [
    1=>12000,
    2=>500,
    3=>11000,
    5=>1000,
    6=>11000,
    8=>12000,
    9=>12000,
    11=>11000,
    13=>12000,
    25=>22000,
    65=>123123
];
$totals = 0;
foreach($uids as $k => $uid){
    $amont = $amounts[$uid];
    if($amont >= $boundary){
        $totals += $amont;
    }else{
        $next = get_next($uids ,$k+1 ,$amont);
        if($next && is_array($next)){
            $curKey = $next[0]; //amouts index 3
            $totals+=$next[2];
            //再向下获取一层
            $nextKey = $curKey+1;
            if(!isset($uids[$nextKey])){
                break;
            }
            $nextUid = $uids[$nextKey];
            $nextAmount = $amounts[$nextUid];
            if($nextAmount >= $boundary){
                $totals+=$nextAmount;
            }else{
                $last = get_next($uids ,$nextKey+1 ,$nextAmount);
                if($last && is_array($last)){
                    $totals+=$last[2];
                }
            }
        }
        break; //跳出主循环
    }
}
echo $totals;
exit;

function get_next($uids ,$start ,$prevAmount){
    global $amounts ,$boundary;
    $leaves = array_slice($uids ,$start ,count($uids),true);
    if($leaves){
        foreach($leaves as $k=>$uid){
            $amount = $prevAmount+$amounts[$uid];
            if($amount >= $boundary){
                return [$k ,$uid ,$amount];
                break;
            }else{
                return get_next($uids ,$k+1 ,$amount);
            }
        }
    }
    return 0;
}</code>

得出$totals=47500

回覆內容:

寫的不好,請大家優化
一個用戶ID的陣列

<code>$uids = [1,2,3,5,6,8,9,11,13,25,65];</code>

這個陣列的每一個鍵值代表一個UID
金額數組

<code>$amounts = [
    1=>12000,
    2=>500,
    3=>11000,
    5=>1000,
    6=>11000,
    8=>12000,
    9=>12000,
    11=>11000,
    13=>12000,
    25=>22000,
    65=>123123
];</code>

此數組的鍵名和$uid數組的鍵值對應。
循環$uids這個數組,取出對應的金額$amount,
如果這個金額大於等於12000($boundary),就在總金額($totals)加上這個$amount
如果這個金額小於12000,就再向下循環,循環到這幾個$amount相加大於等於12000,然後在總金額($totals)加上這幾個$amount的和
最多取三層。
最後得出$totals的值。
我現在的程式碼:

<code><?php
$boundary = 12000;
$uids= [1,2,3,5,6,8,9,11,13,25,65];
$amounts = [
    1=>12000,
    2=>500,
    3=>11000,
    5=>1000,
    6=>11000,
    8=>12000,
    9=>12000,
    11=>11000,
    13=>12000,
    25=>22000,
    65=>123123
];
$totals = 0;
foreach($uids as $k => $uid){
    $amont = $amounts[$uid];
    if($amont >= $boundary){
        $totals += $amont;
    }else{
        $next = get_next($uids ,$k+1 ,$amont);
        if($next && is_array($next)){
            $curKey = $next[0]; //amouts index 3
            $totals+=$next[2];
            //再向下获取一层
            $nextKey = $curKey+1;
            if(!isset($uids[$nextKey])){
                break;
            }
            $nextUid = $uids[$nextKey];
            $nextAmount = $amounts[$nextUid];
            if($nextAmount >= $boundary){
                $totals+=$nextAmount;
            }else{
                $last = get_next($uids ,$nextKey+1 ,$nextAmount);
                if($last && is_array($last)){
                    $totals+=$last[2];
                }
            }
        }
        break; //跳出主循环
    }
}
echo $totals;
exit;

function get_next($uids ,$start ,$prevAmount){
    global $amounts ,$boundary;
    $leaves = array_slice($uids ,$start ,count($uids),true);
    if($leaves){
        foreach($leaves as $k=>$uid){
            $amount = $prevAmount+$amounts[$uid];
            if($amount >= $boundary){
                return [$k ,$uid ,$amount];
                break;
            }else{
                return get_next($uids ,$k+1 ,$amount);
            }
        }
    }
    return 0;
}</code>

得出$totals=47500

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