如何使用数字反转字符串，但不要反转1和0？

Question

I am learning random algorithms, and I am currently stock in one, where I have to reverse a string that contains numbers, but I am not to reverse 1 and 0 in the string eg, 2345678910 would be 1098765432.

Here's what I've done so far:

 function split(str) { let temp = []; temp = str.split(''); const backwards = []; const totalItems = str.length - 1; for (let i = totalItems; i >= 0; i--) { backwards.push(temp[i]); } return backwards.join('').toString(); } console.log(split("10 2 3 USA")); console.log(split("2345678910")); 

I am currently having the issue of not reversing the 10.

What am I doing wrong?

Answer 1

You can replace 10 with a specified character which does not exist in the text, and after running the implemented algorithm replace it back with 10 .

 let out_of_alphabet_character = '#'; var reg_for_the_alphabet = new RegExp(out_of_alphabet_character, "g"); function specific_revert(str) { str = str.replace(/(10)/g, out_of_alphabet_character); let temp = []; temp = str.split(''); const backwards = []; const totalItems = str.length - 1; for (let i = totalItems; i >= 0; i--) { backwards.push(temp[i]); } return backwards.join('').toString().replace(reg_for_the_alphabet, '10'); } console.log(specific_revert("10 2 3 USA")); console.log(specific_revert("234567891010")); 

Answer 2

You can reduce over the matched array from using a regular expression. It's more costly than a for/loop that concatenates strings, but it was fun figuring it out.

 function split(str) { const re = /([A-Z23456789 ]+)|(10)/g return str.match(re).reduce((acc, c) => { // if the match is 10 prepend it to the accumulator // otherwise reverse the match and then prepend it acc.unshift(c === '10' ? c : [...c].reverse().join('')); return acc; }, []).join(''); } console.log(split('2345678910')); console.log(split('10 2 3 US A')); console.log(split('2 3 US A10')); 

Answer 3

Just check for the special case & code the normal logic or reversing as usual

  const reverse = str => { let rev = ""; for (let i = 0; i < str.length; i++) { if (str[i] === '1' && i + 1 < str.length && str[i+1] === '0') { rev = '10' + rev; i++; } else rev = str[i] + rev; } return rev; } console.log(reverse("10 2 3 USA")); // returns ASU 3 2 10 console.log(reverse("2345678910")); // returns 1098765432 

Answer 4

You need some pre-conditions to check each character's value.

Due to the vagueness of the question, it is reasonable to believe that the number system that OP defines consists of [2, 3, 4, 5, 6, 7, 8, 9, 10] and all other characters AZ (including 0 and 1) are simply characters.

 String.prototype.isNumeric = function() { return !isNaN(parseFloat(this)) && isFinite(this); }; function reverse(str) { let tokens = [], len = str.length; while (len--) { let char = str.charAt(len); if (char.isNumeric()) { if (len > 0 && str.charAt(len - 1).isNumeric()) { let curr = parseInt(char, 10), next = parseInt(str.charAt(len - 1), 10); if (curr === 0 && next === 1) { tokens.push(10); len--; continue; } } } tokens.push(char); } return tokens.join(''); } console.log(reverse("10 2 3 USA")); console.log(reverse('2345678910')); 

Output:

 ASU 3 2 10 1098765432 

Answer 5

Below is a recursive approach.

 function f(s, i=0){ if (i == s.length) return ''; if (['0', '1'].includes(s[i])){ let curr = s[i]; while (['0', '1'].includes(s[++i])) curr += s[i] return f(s, i) + curr; } return f(s, i + 1) + s[i]; } console.log(f('10 2 3 US A')); console.log(f('2345678910')); console.log(f('USA101001')); 

Answer 6

Nice question so far.

You may try this recursive approach(if not changing 10 for other character not allowed):

 function reverseKeepTen(str, arr = []) { const tenIdx = str.indexOf('10'); if (!str.length) { return arr.join(''); } if (tenIdx === -1) { return [...str.split('').reverse(), ...arr].join(''); } else { const digitsBefore = str.slice(0, tenIdx); const arrBefore = digitsBefore ? [...digitsBefore.split(''), 10].reverse() : [10]; return reverseKeepTen(str.slice(tenIdx + 2), [...arrBefore, ...arr]) } }; console.log(reverseKeepTen('101234105678910')) // 109876510432110 console.log(reverseKeepTen('12341056789')) // 98765104321 console.log(reverseKeepTen('1012345')) // 5432110 console.log(reverseKeepTen('5678910')) // 1098765 console.log(reverseKeepTen('10111101')) // 11011110