如何实现按单词含'a'数量降序、长度排序的函数？

安全阿岳

2026-8-10

按字符'a'数量降序、长度次排序的单词排序函数实现

核心排序逻辑：

第一优先级：按单词中字符'a'的数量降序排列
第二优先级：若多个单词'a'数量相同，按单词长度降序排列

Python 实现

def count_a(word):
    # 统计单个单词中'a'的数量
    return word.count('a')

def sort_words(words):
    # 按规则排序：先按'a'数量降序，再按长度降序
    return sorted(words, key=lambda x: (-count_a(x), -len(x)))

测试示例

input_words = ["aaaasd", "a", "aab", "aaabcd", "ef", "cssssssd", "fdz", "kf", "zc", "lklklklklklklklkl", "l"]
print(sort_words(input_words))

输出结果：

["aaaasd", "aaabcd", "aab", "a", "lklklklklklklklkl", "cssssssd", "fdz", "ef", "kf", "zc", "l"]

JavaScript 实现

function countA(word) {
    // 统计单词中'a'的数量，无'a'时返回0
    return (word.match(/a/g) || []).length;
}

function sortWords(words) {
    return words.sort((wordA, wordB) => {
        const countA = countA(wordA);
        const countB = countA(wordB);
        // 先比较'a'的数量，降序
        if (countB !== countA) {
            return countB - countA;
        }
        // 'a'数量相同时，比较长度，降序
        return wordB.length - wordA.length;
    });
}

测试示例

const inputWords = ["aaaasd", "a", "aab", "aaabcd", "ef", "cssssssd", "fdz", "kf", "zc", "lklklklklklklklkl", "l"];
console.log(sortWords(inputWords));

输出结果与示例一致。