Lintcode：字符串压缩

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问题：

设计一种方法，通过给重复字符计数来进行基本的字符串压缩。

例如，字符串 aabcccccaaa 可压缩为 a2b1c5a3 。而如果压缩后的字符数不小于原始的字符数，则返回原始的字符串。

可以假设字符串仅包括 a-z 的字母。

样例：

样例 1：

输入：str = "aabcccccaaa"
输出："a2b1c5a3"

样例 2：

输入：str = "aabbcc"
输出："aabbcc"

python：

class Solution:
    """
    @param originalString: a string
    @return: a compressed string
    """
    def compress(self, originalString):
        # write your code here
        if originalString == None or len(originalString) <= 1:
            return originalString
        result = ""
        result += originalString[0]
        strIndex = 1
        for i in range(1, len(originalString)):
            if originalString[i] == originalString[i-1]:
                strIndex += 1
            else:
                result += str(strIndex)
                result += originalString[i]
                strIndex = 1
        result += str(strIndex)
        if len(result) > len(originalString):
            return originalString
        for j in range(1, len(result),2):
            temp = int(result[j])
            if  temp >= 3:
                return result
        return originalString

C++：

class Solution {
public:
    /**
     * @param originalString: a string
     * @return: a compressed string
     */
    string compress(string &originalString) {
        // write your code here
        if(originalString == "" || originalString.size() == 1)
        {
            return originalString;
        }
        string result = "";
        result += originalString[0];
        int strIndex = 1;
        for(int i = 1; i < originalString.size(); i++)
        {
            if(originalString[i] == originalString[i-1])
            {
                strIndex++;
            }else{
                result += to_string(strIndex);
                result += originalString[i];
                strIndex = 1;
            }
        }
        result += to_string(strIndex);
        if(result.size() > originalString.size())
        {
            return originalString;
        }
        for(int j = 1; j < result.size(); j+=2)
        {
            int temp = int(result[j] - '0');
            if(temp >= 3)
            {
                return result;
            }
        }
        return originalString;
    }
};