PHP看上去-说顺序

Question

我试图在PHP中编写Conway外观序列代码。这是我的代码：

function look_and_say ($number) {
    $arr = str_split($number . " ");
    $target = $arr[0];
    $count = 0;
    $res = "";
    foreach($arr as $num){
        if($num == $target){
            $count++;
        }else{
            $res .= $count . $target;
            $count = 1;
            $target = $num;
        }        
    }
    return $res;
}

当我运行这个函数时，look_and_say(9900)得到了我所期望的值：2920。我的问题是将$arr指定为$arr = str_split($number)而不是$arr = str_split($number . " ")，结果忽略了$arr的最后一个元素并返回29。

在$arr前端末尾添加空空间以检查最后一个元素是正常的，还是有更好的方法来实践这段代码--除了regex方式。

Answer 1

我想出了两种方法。

1也是在循环后的结果中添加级联。

function look_and_say ($number) {
    $arr = str_split($number);
    $target = $arr[0];
    $count = 0;
    $res = "";
    foreach($arr as $num){
        if($num == $target){
            $count++;
        }else{
            $res .= $count . $target;
            $count = 1;
            $target = $num;
        }        
    }
    $res .= $count . $target;
    return $res;
}

第二个是在循环中添加另一个if子句并确定最后一个迭代：

function look_and_say ($number) {
    $arr = str_split($number);
    $target = $arr[0];
    $count = 0;
    $res = "";
    $i=0;
    $total = count($arr);
    foreach($arr as $num){        
        if($i == ($total-1))
        {
            $count++;
            $res .= $count . $target;
        }
        elseif($num == $target){
            $count++;
        }else{
            $res .= $count . $target;
            $count = 1;
            $target = $num;
        }
        $i++;
    }
    return $res;
}

Answer 2

我建议您以其他方式使用两个嵌套的while循环：

<?php

function lookAndSay($number) {
    $digits = str_split($number);
    $result = '';
    $i = 0;
    while ($i < count($digits)) {
        $lastDigit = $digits[$i];
        $count = 0;
        while ($i < count($digits) && $lastDigit === $digits[$i]) {
            $i++;
            $count++;
        }
        $result .= $count . $lastDigit;
    }

    return $result; 
}