示例#1
0
<?php

function isAnagram($satu, $dua)
{
    $satuAr = str_split($satu);
    $duaAr = str_split($dua);
    $jumlah = 0;
    foreach ($satuAr as $k => $v) {
        if (!in_array($v, $duaAr)) {
            $jumlah++;
        }
    }
    if ($jumlah != 0) {
        echo "false";
    } else {
        echo "true";
    }
}
#isAnagram("secure","resc ue");
isAnagram("conifers", "fir cones");
#isAnagram("google","facebook");
示例#2
0
<?php

$wordPairs[] = array("cat", "act");
// happy path
$wordPairs[] = array("cat", "apt");
// not so happy path
$wordPairs[] = array("cat", "pact");
// not so happy path
$wordPairs[] = array("cat", "tact");
// not so happy path
$wordPairs[] = array("31121984", "12311984");
// happy path
foreach ($wordPairs as $words) {
    if (isAnagram($words[0], $words[1])) {
        echo "{$words['0']}, {$words['1']} are anagrams. \n";
    } else {
        echo "{$words['0']}, {$words['1']} are not anagrams. \n";
    }
}
function isAnagram($word1, $word2)
{
    if (empty($word1) || empty($word1) || strlen($word1) != strlen($word2)) {
        // check for empty strings or if
        // one of the words atleast has one or more chars more than the other
        return false;
    } else {
        $charFreqMapWord1 = __buildFrequencyMap($word1);
        print_r($charFreqMapWord1);
        $charFreqMapWord2 = __buildFrequencyMap($word2);
        print_r($charFreqMapWord2);
        foreach ($charFreqMapWord1 as $char => $count) {