Lintcode 213: Sring Compression

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本文介绍了一种基本的字符串压缩方法，使用字符重复计数进行压缩。例如，字符串'aabcccccaaa'将被压缩为'a2b1c5a3'。如果压缩后的字符串长度不小于原字符串，则返回原字符串。文章提供了详细的算法实现，并附带了示例。

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String Compression
Implement a method to perform basic string compression using the counts of repeated characters. For example, the string aabcccccaaa would become a2b1c5a3.

If the “compressed” string would not become smaller than the original string, your method should return the original string.

You can assume the string has only upper and lower case letters (a-z).

Example
str=aabcccccaaa return a2b1c5a3
str=aabbcc return aabbcc
str=aaaa return a4

解法：
1）注意用to_string()将counters[currChar - ‘a’]换成string

class Solution {
public:
    /**
     * @param originalString: a string
     * @return: a compressed string
     */
    string compress(string &originalString) {
        vector<int> counters(26, 0);   //counter of each letter is it is in series
        string newString;
        int len = originalString.size();
        
        char currChar = originalString[0];
        counters[currChar  - 'a'] = 1;
        
        for (int i = 1; i < len; ++i) {
            if (originalString[i] != currChar) {
                newString = newString + currChar + to_string(counters[currChar - 'a']);
                counters[currChar - 'a'] = 0;
                currChar = originalString[i];
            }
            counters[currChar - 'a']++; 
        }
        
        newString = newString + currChar + to_string(counters[currChar - 'a']);
        
        if (newString.size() >= len) return originalString;
        else return newString;
    }
};