[LeetCode] 482. License Key Formatting 注册码格式化

You are given a license key represented as a string S which consists only alphanumeric character and dashes. The string is separated into N+1 groups by N dashes.

Given a number K, we would want to reformat the strings 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 all lowercase letters should be converted to uppercase.

Given a non-empty string S and a number K, format the string according to the rules described above.

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.

Note:

1. The length of string S will not exceed 12,000, and K is a positive integer.
2. String S consists only of alphanumerical characters (a-z and/or A-Z and/or 0-9) and dashes(-).
3. String S is non-empty.

给一个字符串的注册码S，按要求对注册码进行格式化，每四个字符后面跟一个短横杠，每一部分的长度为K，第一部分长度可以小于K，所有字母必须是大写的。

解法：用数组记录字符，从后往前循环字符串，当前字符不是短横杠，就变大写加入数组，每隔四个字母加一个短横杠，最后在转换成字符串。

最重要的是判断4个字符的方法： if len(res) % (K + 1) == K

Java:

class Solution {
    public String licenseKeyFormatting(String S, int K) {
        StringBuilder sb = new StringBuilder();
        for (int i = S.length() - 1; i >= 0; i--) {
            if (S.charAt(i) != '-') {
                sb.append(sb.length() % (K + 1) == K ? "-" : "").append(S.charAt(i));
            } 
        }
        return sb.reverse().toString().toUpperCase();
    }
} 

Java:

public String licenseKeyFormatting(String s, int k) {
    StringBuilder sb = new StringBuilder();
    for (int i = s.length() - 1; i >= 0; i--)
        if (s.charAt(i) != '-')
            sb.append(sb.length() % (k + 1) == k ? '-' : "").append(s.charAt(i));
    return sb.reverse().toString().toUpperCase();
} 　　

Python: wo

class Solution(object):
    def licenseKeyFormatting(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        res = []
        for i in reversed(range(len(S))):
            if S[i] != '-':
                if len(res) % (K + 1) == K:
                    res.append('-')
                res.append(S[i].upper())
                
        return ''.join(s for s in reversed(res))   　

Python: wo

class Solution(object):
    def licenseKeyFormatting(self, S, K):
        res = []
        for c in reversed(S):
            if c != '-':
                if len(res) % (K + 1) == K:
                    res.append('-')
                res.append(c.upper())
                
        return ''.join(r for r in reversed(res)) 　　

Python:

class Solution(object):
    def licenseKeyFormatting(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        S = S.upper().replace('-','')
        size = len(S)
        s1 = K if size%K==0 else size%K
        res = S[:s1]
        while s1<size:
            res += '-'+S[s1:s1+K]
            s1 += K
        return res　

Python:

class Solution:
    def licenseKeyFormatting(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        S = S.replace("-", "").upper()[::-1]
        return '-'.join(S[i:i+K] for i in range(0, len(S), K))[::-1]　　

Python:

class Solution:
    def licenseKeyFormatting(self, S, K):
        """
        :type S: str
        :type K: int
        :rtype: str
        """
        def chunks(l, n):
            for i in range(0, len(l), n):
                yield l[i:i+n]
        s = S[::-1].upper().replace('-', '')   
        return '-'.join(list(chunks(s, K)))[::-1]　　

C++:

class Solution {
public:
    string licenseKeyFormatting(string S, int K) {
        string res = "";
        for (int i = (int)S.size() - 1; i >= 0; --i) {
            if (S[i] != '-') {
                ((res.size() % (K + 1) - K) ? res : res += '-') += toupper(S[i]);
            }
        }
        return string(res.rbegin(), res.rend());
    }
};

　　

 

 

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转载于:https://www.cnblogs.com/lightwindy/p/9747402.html