# [443. String Compression](https://leetcode.com/problems/string-compression/description/)
# 思路
根据题意压缩字符串。  
用res记录当前处理过的字符串压缩后的长度，num记录当前字符出现了多少次。
# C++
```
class Solution {
public:
    int compress(vector<char>& chars) {
        int res = 0, tmp, num = 1; // num初始为1
        for(int i = 1; i < chars.size(); i++){
            if(chars[i] == chars[i - 1]) num++;
            else{
                chars[res++] = chars[i - num];
                if(num == 1) continue; // 若出现次数仅为1则后面不加“1”
                
                // 以下5行将int型转为char型，也可用string s = to_string(num)转，不过好像慢一些
                tmp = res;
                while(num != 0){
                    chars[res++] = (num % 10 + '0');
                    num /= 10;
                }
                reverse(chars.begin() + tmp, chars.begin() + res);
                num = 1;
            }
        }
        chars[res++] = chars[chars.size() - 1];
        if(num > 1){
            tmp = res;
            while(num != 0){
                chars[res++] = (num % 10 + '0');
                num /= 10;
            }
            reverse(chars.begin() + tmp, chars.begin() + res);
        }        
        return res;
    }
};
```