443. String Compression

Given an array of characters, compress it in-place.

The length after compression must always be smaller than or equal to the original array.

Every element of the array should be a character (not int) of length 1.

After you are done modifying the input array in-place, return the new length of the array.

Follow up:
Could you solve it using only O(1) extra space?

Example 1:

Input:
[“a”,”a”,”b”,”b”,”c”,”c”,”c”]
Output:
Return 6, and the first 6 characters of the input array should be: [“a”,”2”,”b”,”2”,”c”,”3”]
Explanation:
“aa” is replaced by “a2”. “bb” is replaced by “b2”. “ccc” is replaced by “c3”.

Example 2:

Input:
[“a”]
Output:
Return 1, and the first 1 characters of the input array should be: [“a”]
Explanation:
Nothing is replaced.

Example 3:

Input:
[“a”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”,”b”]
Output:
Return 4, and the first 4 characters of the input array should be: [“a”,”b”,”1”,”2”].
Explanation:
Since the character “a” does not repeat, it is not compressed. “bbbbbbbbbbbb” is replaced by “b12”.
Notice each digit has it’s own entry in the array.

Note:
1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

注意是in-place,不可再开辟存储空间，在chars内完成操作

class Solution {
public:
    int compress(vector<char>& chars) {
        if(chars.size()==0)
            return 0;
        int ans=0,index=0;
        while(index<chars.size())
        {
            char c=chars[index];
            int count=0;
            while(index<chars.size() && chars[index]==c)
            {
                index++;
                count++;
            }
            chars[ans++]=c;
            if(count!=1)
            {
                for(char a:to_string(count))
                    chars[ans++]=a;
            }
        }
        return ans;      
    }
};