Create 443. String Compression.md

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# [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;
}
};
```