javascript - 创建一个接受字符串并将重复值分组的函数

Question

创建一个接受字符串并对重复值进行分组的函数。这些组应具有以下结构:[[value, first_index, last_index, times_repeated], ..., [value, first_index, last_index, times_repeated]]。

• 值:被评估的字符。
• first_index:字符首次出现的索引。
• last_index:字符最后一次出现的索引。
• times_repeated:字符重复的连续次数。

例子

findRepeating("a") ➞ [["a", 0, 0, 1]]

findRepeating("aabbb") ➞ [["a", 0, 1, 2], ["b", 2, 4, 3]]

findRepeating("1337") ➞ [["1", 0, 0, 1], ["3", 1, 2, 2], ["7", 3, 3, 1]]

findRepeating("aabbbaabbb") ➞ [["a", 0, 1, 2], ["b", 2, 4, 3], ["a", 5, 6, 2], ["b", 7, 9, 3]]

我能够为独特的 Angular 色做到这一点。但无法为

字符重复的连续次数

我的代码

function findRepeating(str) {
    let unique = [...new Set([...str])]
    return unique.map(x=>[x,str.indexOf(x),str.lastIndexOf(x),[...str].filter(a=>a==x).length])
}

预期结果

Test.assertSimilar(findRepeating(''), [])

Test.assertSimilar(findRepeating('a'), [['a', 0, 0, 1]])

Test.assertSimilar(findRepeating('1337'), [['1', 0, 0, 1], ['3', 1, 2, 2], ['7', 3, 3, 1]])

Test.assertSimilar(findRepeating('aabbb'), [['a', 0, 1, 2], ['b', 2, 4, 3]])

Test.assertSimilar(findRepeating('addressee'), [['a', 0, 0, 1], ['d', 1, 2, 2], ['r', 3, 3, 1], ['e', 4, 4, 1], ['s', 5, 6, 2], ['e', 7, 8, 2]])

Test.assertSimilar(findRepeating('aabbbaabbb'), [['a', 0, 1, 2], ['b', 2, 4, 3], ['a', 5, 6, 2], ['b', 7, 9, 3]])

Test.assertSimilar(findRepeating('1111222233334444'), [['1', 0, 3, 4], ['2', 4, 7, 4], ['3', 8, 11, 4], ['4', 12, 15, 4]])

Test.assertSimilar(findRepeating('1000000000000066600000000000001'), [['1', 0, 0, 1], ['0', 1, 13, 13], ['6', 14, 16, 3], ['0', 17, 29, 13], ['1', 30, 30, 1]])

实际结果

Test Passed: Value == '[]'

Test Passed: Value == "[['a', 0, 0, 1]]"

Test Passed: Value == "[['1', 0, 0, 1], ['3', 1, 2, 2], ['7', 3, 3, 1]]"

Test Passed: Value == "[['a', 0, 1, 2], ['b', 2, 4, 3]]"

FAILED: Expected: "[['a', 0, 0, 1], ['d', 1, 2, 2], ['r', 3, 3, 1], ['e', 4, 4, 1], ['s', 5, 6, 2], ['e', 7, 8, 2]]", instead got: "[['a', 0, 0, 1], ['d', 1, 2, 2], ['r', 3, 3, 1], ['e', 4, 8, 3], ['s', 5, 6, 2]]"

FAILED: Expected: "[['a', 0, 1, 2], ['b', 2, 4, 3], ['a', 5, 6, 2], ['b', 7, 9, 3]]", instead got: "[['a', 0, 6, 4], ['b', 2, 9, 6]]"

Test Passed: Value == "[['1', 0, 3, 4], ['2', 4, 7, 4], ['3', 8, 11, 4], ['4', 12, 15, 4]]"

FAILED: Expected: "[['1', 0, 0, 1], ['0', 1, 13, 13], ['6', 14, 16, 3], ['0', 17, 29, 13], ['1', 30, 30, 1]]", instead got: "[['1', 0, 30, 2], ['0', 1, 29, 26], ['6', 14, 16, 3]]"

function findRepeating(str) {
	let unique = [...new Set([...str])]
	return unique.map(x=>[x,str.indexOf(x),str.lastIndexOf(x),[...str].filter(a=>a==x).length])
}
console.log("Fails   ",JSON.stringify(findRepeating('addressee')),"\nexpected", `[['a',0,0,1],['d',1,2,2],['r',3,3,1],['e',4,4,1],['s',5,6,2],['e',7,8,2]]`)
console.log("Fails   ",JSON.stringify(findRepeating('aabbbaabbb')),"\nexpected", `[['a',0,1,2],['b',2,4,3],['a',5,6,2],['b',7,9,3]]`)
console.log("Passes  ",JSON.stringify(findRepeating('1111222233334444')),"\nexpected", `[['1',0,3,4],['2',4,7,4],['3',8,11,4],['4',12,15,4]]`)
console.log("Fails   ",JSON.stringify(findRepeating('1000000000000066600000000000001')),"\nexpected", `[['1',0,0,1],['0',1,13,13],['6', 14,16,3],['0',17,29,13],['1',30,30,1]]`)

Answer 1

您可以使用正则表达式获取一组相同字符，该正则表达式查找一个字符和相同的后续一次作为一个组并映射所需信息。

function findRepeating(string) {
    var i = -1;
    return (string.match(/(.)\1*/g) || []).map(s => [s[0], ++i, i += s.length - 1, s.length]);
}

console.log(findRepeating(""));           // []
console.log(findRepeating("a"));          // [["a", 0, 0, 1]]
console.log(findRepeating("aabbb"));      // [["a", 0, 1, 2], ["b", 2, 4, 3]]
console.log(findRepeating("1337"));       // [["1", 0, 0, 1], ["3", 1, 2, 2], ["7", 3, 3, 1]]
console.log(findRepeating("aabbbaabbb")); // [["a", 0, 1, 2], ["b", 2, 4, 3], ["a", 5, 6, 2], ["b", 7, 9, 3]]