计算常用字符的频率

接下来,找到每个字符串之间的公共字符,并创建一个包含每个公共字符频率的映射.

这里最大的性能提升是创建一个独特的字符集中出现的所有字符集.我利用Set对象的独特性来从数组中删除重复项,还有其他方法可以从数组中删除重复项,但这是我测试过的最快的重复项.这样,我们只迭代一个unqiue集并检查两个映射中是否出现该字符.

// Calculate the frequency of each character two strings have in common
const calculateCommonCharacterFrequency = (v1, v2) => {
    // Get frequency map of first string
    v1 = calculateCharacterFrequency(v1);
    // Get frequency map of second string
    v2 = calculateCharacterFrequency(v2);
    // Create an array containing a list of all characters occuring in either string
    const combinedCharacterArray = [...v1.keys(), ...v2.keys()];
    // Remove duplicate characters
    const combinedUniqueCharacterSet = new Set(combinedCharacters);
    // Convert set back to array
    const combinedUniqueCharacterArray = Array.from(combinedUniqueSet);
    // Iterate the unique array of common characters
    const commonFrequency = combinedUniqueCharacterArray.reduce(
        (commonFrequency, character) => {
            // Check if both strings have the character
            const isCommon = v1.has(character) && v2.has(character);
            if(isCommon) {
                // Find the sum of the frequency of the character in both strings
                const combinedFrequency = v1.get(character) + v2.get(character);
                // Insert character and frequency into map
                commonFrequency.set(character, combinedFrequency);
            }
            // return frequency map for use in next iteration
            return commonFrequency;
        }, 
        // This is the map that will be passed as `commonFrequency`
        new Map()
    );
    // Return the map
    return commonFrequency;
};

const calcCharFreq = string => [...string].sort().join("").match(/(.)\1*/g).reduce((freq, char) => (freq.set(char[0], char.length), freq), new Map());

const calcCommonCharFreq = (v1, v2) => {
    v1 = calcCharFreq(v1);
    v2 = calcCharFreq(v2);
    return Array.from(new Set([...v1.keys(), ...v2.keys()])).reduce((dup, char) => ((v1.has(char) && v2.has(char)) && dup.set(char, v1.get(char) + v2.get(char)), dup), new Map());
};

const test = (v1,v2) => (console.log('{ '+Array.from(calcCommonCharFreq(v1, v2)).reduce((pairs, value) => (pairs.push('"'+value[0]+'": '+value[1]), pairs), []).join(", ")+' }'), test);

test("Leopard", "Lion")            // { "L": 2, "o": 2 }
    ("Hello", "World")             // { "l": 3, "o": 2 }
    ("Flower", "Power")            // { "o": 2, "w": 2, "e": 2, "r": 2 }
    ("phone", "Memphis")           // { "p": 2, "h": 2, "e": 2 }
    ("Bookkeeper", "Zookeeper")    // { "o": 4, "k": 3, "e": 6, "p": 2, "r": 2 }
    ("Some thing", "Other thing"); // { "e": 2, " ": 2, "t": 3, "h": 3, "i": 2, "n": 2, "g": 2 }

const countLetters = (v1, v2) => {
  const out = {};
  
  for(let char of v1) (out[char] || v2.indexOf(char) !== -1) && (out[char] = (out[char] || 0) + 1);
  for(let char of v2) (out[char] || v1.indexOf(char) !== -1) && (out[char] = (out[char] || 0) + 1);

  return out;
}

const test = (v1,v2) => (console.log(JSON.stringify(countLetters(v1, v2))),test);

test("Leopard", "Lion")            // { "L": 2, "o": 2 }
    ("Hello", "World")             // { "l": 3, "o": 2 }
    ("Flower", "Power")            // { "o": 2, "w": 2, "e": 2, "r": 2 }
    ("phone", "Memphis")           // { "p": 2, "h": 2, "e": 2 }
    ("Bookkeeper", "Zookeeper")    // { "o": 4, "k": 3, "e": 6, "p": 2, "r": 2 }
    ("Some thing", "Other thing"); // { "e": 2, " ": 2, "t": 3, "h": 3, "i": 2, "n": 2, "g": 2 }