1417. Reformat The String

JIAKAOBO

LeetCode

感谢赞助!

• ㊗️
• 大家
• offer
• 多多！

Problem

Given alphanumeric string s. (Alphanumeric string is a string consisting of lowercase English letters and digits).

You have to find a permutation of the string where no letter is followed by another letter and no digit is followed by another digit. That is, no two adjacent characters have the same type.

Return the reformatted string or return an empty string if it is impossible to reformat the string.

Example 1:

Input: s = "a0b1c2"
Output: "0a1b2c"
Explanation: No two adjacent characters have the same type in "0a1b2c". "a0b1c2", "0a1b2c", "0c2a1b" are also valid permutations.

Example 2:

Input: s = "leetcode"
Output: ""
Explanation: "leetcode" has only characters so we cannot separate them by digits.

Example 3:

Input: s = "1229857369"
Output: ""
Explanation: "1229857369" has only digits so we cannot separate them by characters.

Example 4:

Input: s = "covid2019"
Output: "c2o0v1i9d"

Example 5:

Input: s = "ab123"
Output: "1a2b3"

Code

class Solution {
    public String reformat(String s) {
        StringBuilder letter = new StringBuilder();
        StringBuilder digit = new StringBuilder();

        for(char c : s.toCharArray()) {
            if(Character.isDigit(c)) {
                digit.append(c);
            } else {
                letter.append(c);
            }
        }

        int len1 = letter.length();
        int len2 = digit.length();

        if(len1 != len2 &&
        Math.abs(len1 - len2) != 1) return "";

        StringBuilder res = new StringBuilder();

        int i = 0;
        int j = 0;

        boolean isLetter = true;
        if(len1 > len2) {
            res.append(letter.charAt(i++));
            isLetter = false;
        } else if (len1 < len2) {
            res.append(digit.charAt(j++));
            isLetter = true;
        }

        while(i < len1 || j < len2) {
            if(isLetter == true) {
                res.append(letter.charAt(i++));
                res.append(digit.charAt(j++));
            } else {
                res.append(digit.charAt(j++));
                res.append(letter.charAt(i++));
            }
        }
        return res.toString();
    }
}