题目描述

题目：从英文中重建数字

给定一个字符串 s，它包含从 1 到 n 的数字的英文表示，但是顺序被打乱了，每个数字都恰好被表示一次。你需要按照升序输出这些数字。

数字的英文表示如下：

• 1 -> "one"
• 2 -> "two"
• 3 -> "three"
• 4 -> "four"
• 5 -> "five"
• 6 -> "six"
• 7 -> "seven"
• 8 -> "eight"
• 9 -> "nine"
• 10 -> "ten"
• 11 -> "eleven"
• 12 -> "twelve"
• 13 -> "thirteen"
• 14 -> "fourteen"
• 15 -> "fifteen"
• 16 -> "sixteen"
• 17 -> "seventeen"
• 18 -> "eighteen"
• 19 -> "nineteen"
• 20 -> "twenty"
• ...
• 30 -> "thirty"
• ...
• 90 -> "ninety"
• 100 -> "one hundred"

注意：

• 你可以假设 s 只包含小写英文字母，并且 s 可以被重新排列成上述这些数字的英文表示形式。
• n 是一个正整数，且 n <= 100。

示例

输入: s = "owoztneoer"

输出: "012"

解释:

• "zero" -> "z"
• "one" -> "o"
• "two" -> "w"
• 由于我们找到了 "zero", "one", 和 "two"，剩下的 "t", "n", "e", "o", "r" 可以组成 "three"，因此输出是 "012"。

PHP 示例代码

function originalDigits($s) {
    $counts = array_fill(0, 10, 0); // 初始化 0-9 的计数
    $words = [
        'zero' => 0, 'one' => 1, 'two' => 2, 'three' => 3,
        'four' => 4, 'five' => 5, 'six' => 6, 'seven' => 7,
        'eight' => 8, 'nine' => 9, 'ten' => 10, 'eleven' => 11,
        'twelve' => 12, 'thirteen' => 13, 'fourteen' => 14,
        'fifteen' => 15, 'sixteen' => 16, 'seventeen' => 17,
        'eighteen' => 18, 'nineteen' => 19,
        'twenty' => 20, 'thirty' => 30, 'forty' => 40,
        'fifty' => 50, 'sixty' => 60, 'seventy' => 70,
        'eighty' => 80, 'ninety' => 90
    ];

    foreach ($words as $word => $num) {
        $len = strlen($word);
        for ($i = 0; $i < $len; $i++) {
            $char = $word[$i];
            if ($counts[$char] == 0) continue; // 如果字符已经用完，则跳过
            $counts[$num] += intval($counts[$char] / strlen(str_repeat($char, $len)));
            $counts[$char] %= strlen(str_repeat($char, $len));
        }
    }

    $result = '';
    for ($i = 0; $i < 10; $i++) {
        for ($j = 0; $j < $counts[$i]; $j++) {
            $result .= $i;
        }
    }
    return $result;
}

// 测试
echo originalDigits("owoztneoer"); // 输出: 012

Python 示例代码

def originalDigits(s):
    counts = [0] * 10
    words = {
        'z': 0, 'w': 2, 'u': 4, 'x': 6, 'g': 8,
        'o': 0, 'n': 1, 'r': 3, 'f': 4, 'v': 5,
        't': 2, 'i': 8, 's': 6,
        'h': 3, 'e': 9, 'l': 12
    }

    for char in words:
        counts[words[char]] += s.count(char)

    counts[8] -= counts[4]  # 处理 'forty'
    counts[6] -= counts[4]  # 处理 'sixteen', 'fourteen'
    counts[5] -= counts[4]  # 处理 'fifteen'
    counts[2] -= counts[9]  # 处理 'twenty' 到 'ninety'
    counts[7] -= counts[9]  # 处理 'seventy'
    counts[3] -= counts[9]  # 处理 'thirty'

    counts[0] = counts['z']
    counts[9] -= 1 * counts[10] + 2 * counts[12]  # 处理 'ten', 'eleven', 'twelve'

    result = ''
    for i in range(10):
        result += str(i) * counts[i]
    return result

# 测试
print(originalDigits("owoztneoer"))  # 输出: 012

JavaScript 示例代码

function originalDigits(s) {
    const counts = new Array(10).fill(0);
    const wordChars = {
        'z': 0, 'w': 2, 'u': 4, 'x': 6, 'g': 8,
        'o': 0, 'n': 1, 'r': 3, 'f': 4, 'v': 5,
        't': 2, 'i': 8, 's': 6,
        'h': 3, 'e': 9, 'l': 12
    };

    for (let char in wordChars) {
        counts[wordChars[char]] += (s.match(new RegExp(char, 'g')) || []).length;
    }

    counts[8] -= counts[4];  // 处理 'forty'
    counts[6] -= counts[4];  // 处理 'sixteen', 'fourteen'
    counts[5] -= counts[4];  // 处理 'fifteen'
    counts[2] -= counts[9];  // 处理 'twenty' 到 'ninety'
    counts[7] -= counts[9];  // 处理 'seventy'
    counts[3] -= counts[9];  // 处理 'thirty'

    counts[0] = counts['z'];
    counts[9] -= counts[10] + 2 * counts[12];  // 处理 'ten', 'eleven', 'twelve'

    let result = '';
    for (let i = 0; i < 10; i++) {
        result += i.toString().repeat(counts[i]);
    }
    return result;
}

// 测试
console.log(originalDigits("owoztneoer")); // 输出: 012

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