LeetCode 482. License Key Formatting

题目

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

解析

题干是很好理解的，即为将所给的字符串S，按照K进行分组，分组之间用“-”进行连接。第一个分组数量只要不超过K就保持不变，超过了也进行拆分。
此题的解题思路为：

1. 求得所有字母的长度len，第一个分组的长度，添加连接符后的长度；
2. 对给定字符串和结果字符串遍历，用三个下标分别遍历给定字符串(j)、结果字符串(i)和连接符的位置(dashIndex)。

难点在于结果字符串连接符的个数的求解，通过和给一个分组“凑单”的方式，便可以求到。
另外在遍历过程中，要注意给定字符串S[j]为'-'时，跳过(++j)，否则进行赋值，并且累加dashIndex。当dashIndex==K+1时，置dashIndex=1。

代码

char toUpperChar(char ch);
int keyLength(char* s);

char * licenseKeyFormatting(char * S, int K){
    int totalLen = keyLength(S);
    int firstGroupLen = totalLen % K;
    
    // indicate characters length can be divided by K with no remainder
    if (firstGroupLen == 0)
        firstGroupLen = K;
    
    // add dashes and get length
    // add (K - 1) length group to gather the first group
    // if the first group can divided by K without remainder, so add this group
    // is meaningless, else add the group in order to add one dashes in the result
    totalLen += (totalLen + (K - 1)) / K - 1;
    
    if (totalLen < 0)
        totalLen = 0;
    
    char* formatKey = (char*)malloc((totalLen + 1) * sizeof(char));
    
    formatKey[totalLen] = '\0';
    
    // i is the loop index of the formatKey array
    // j is the loop index of K
    // dashIndex is to find dash position, initialize to K + 1 - firstGroupLen
    // so as to make the D-value for the first group. It will be set to 1 after find
    // the first position of dash.
    int i = 0, j = 0, dashIndex = K + 1 - firstGroupLen;
    
    while (i < totalLen) {
        if (dashIndex == K + 1) {
            formatKey[i++] = '-';
            
            dashIndex = 1;
        } else {
            // if S[j] is equal to '-', then jump to next,
            // else upper S[j] and set to formatKey[i]
            if (S[j] != '-') {
                formatKey[i++] = toUpperChar(S[j]);
                ++dashIndex;
            }
            
            ++j;
        }
    }
    
    return formatKey;
}

int keyLength(char* s) {
    int len = 0;
    
    for (int i = 0; s[i] != '\0'; ++i) {
        if (s[i] != '-')
            ++len;
    }
    
    return len;
}

char toUpperChar(char ch) {
    if (ch >= 'a' && ch <= 'z')
        ch -= ('a' - 'A');
    
    return ch;
}

在以上代码中，结果字符串中连接符个数为(totalLen + (K - 1)) / K - 1，即通过添加一个分组来对第一个分组进行凑单。如果第一个分组可以整除K，则添加(K-1)则无意义，如果不能整除K，则添加此分组是为了与其拼凑，最后再将多加的1除去，因为可能加多了。
另外，dashIndex的初始化是K+1-firstGroupLen，是为了第一个分组的长度，即找到第一个连接符的位置，然后将dashIndex置1，由于判等的是K+1，因为下标从1开始。
这样便解决了。

此题比较考察逻辑思维，建议读者在阅读完之后用例子中的字符串进行一下推导，这样可以更好的理解。