LeetCode 482. License Key Formatting

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Description

You are given a license key represented as a string s that consists of only alphanumeric characters and dashes. The string is separated into n + 1 groups by n dashes. You are also given an integer k.

We want to reformat the string s such that each group contains exactly k characters, except for the first group, which could be shorter than k but still must contain at least one character. Furthermore, there must be a dash inserted between two groups, and you should convert all lowercase letters to uppercase.

Return the reformatted license key.

Example 1:

Input: s = "5F3Z-2e-9-w", k = 4
Output: "5F3Z-2E9W"
Explanation: The string s has been split into two parts, each part has 4 characters.
Note that the two extra dashes are not needed and can be removed.
Example 2:

Input: s = "2-5g-3-J", k = 2
Output: "2-5G-3J"
Explanation: The string s has been split into three parts, each part has 2 characters except the first part as it could be shorter as mentioned above.

Constraints:

1 <= s.length <= 105
s consists of English letters, digits, and dashes '-'.
1 <= k <= 104

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Analysis

Create a new license string by reverse iterating the input license string. The point is when to add dashes to the new license string,

Solution written in Cpp

class Solution {
public:
    string licenseKeyFormatting(string s, int k) {
        int n = s.length();
        int count = 0;
        string result;

        for (int i = n - 1; i >= 0; i--) {
            if (s[i] != '-') {
                if (count == k) {
                    result.push_back('-');
                    count = 0;
                }
                result.push_back(toupper(s[i]));
                count++;
            }
        }

        reverse(result.begin(), result.end());
        return result;
    }
};

Solution written in Rust

impl Solution {
    pub fn license_key_formatting(s: String, k: i32) -> String {
        let mut ret = String::new();
        let mut count = 0;
        s.chars().rev().for_each(|c| {
            if count != k {
                if c != '-' {
                    if count == 0 && ret.len() != 0 {
                        ret.push('-');
                    }

                    ret.push(c.to_uppercase().next().unwrap());
                    count += 1;
                }
            }

            if count == k {
                count = 0;
            }
        });
        ret.chars().rev().collect::<String>()
    }
}

[quinnwencn]

还可以这么做，将字符串的-都去掉然后收集字符成一个vec，然后再将vec的元素按要求拼接-