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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;
+    }
+};
+```