刷完上一道题目字符串压缩，看到leetcode中的443题，基本上与字符串压缩这道题一致。

class Solution {
public:
    int compress(vector<char>& chars) {
        int n = chars.size();
        //location to write and character count
        int loc = 0,cnt = 1;
        if(n < 2) return n;
        for(int i = 1 ;i < n ;i++)
        {
            if(chars[i] == chars[i-1])
            {
                cnt++;
            }else{
                chars[loc++] = chars[i-1];
                if(cnt != 1)
                {
                    string str_cnt = to_string(cnt);
                    for(auto c:str_cnt)
                        chars[loc++] = c;
                }
                cnt = 1;
            }
            //handle abbb and aaab case.
            if(i == n -1)
            {
                chars[loc++] = chars[i];
                if(cnt != 1){
                    string str_cnt = to_string(cnt);
                    for(auto c:str_cnt)
                        chars[loc++] = c;
                }
            }
        }
        return loc;
    }
};

运行效率：

Runtime: 4 ms, faster than 92.64% of C++ online submissions for String Compression.
Memory Usage: 8.8 MB, less than 78.60% of C++ online submissions for String Compression.

看了讨论区大神的解法非常的精炼，AC代码如下:

class Solution {
public:
    int compress(vector<char>& chars) {
        int n = chars.size();
        if(n < 2) return n;
        //i represent the start of a character,loc resprenet where to write
        int i = 0,loc = 0;
        for(int j = 0;j < n;j++)
        {
            if(j == n-1 || chars[j] != chars[j+1])
            {
                chars[loc++] = chars[j];
                //skip length== 1 case
                if(i < j)
                {
                    //j - i + 1 > 1,represent the character count
                    for(auto c:to_string(j - i + 1))
                        chars[loc++] = c;
                }
                //update start position
                i = j + 1;
            }
            
        }
        return loc;
    }
};