题目描述：
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:
- All characters have an ASCII value in
[35, 126]. 1 <= len(chars) <= 1000.
将字符串压缩，用字符加次数的方式表示连续重复的子字符串，而且要求直接在原字符串修改。方法就是遍历字符串，统计当前字符出现的次数，并且用指针标记当前遍历的位置和新字符串的尾部，这样就便于直接在原字符串修改。
class Solution {
public:
int compress(vector<char>& chars) {
if(chars.size()==0) return 0;
int count=0;
int i=0;
int j=0;
char cur=chars[0];
while(j<chars.size())
{
if(cur==chars[j])
{
count++;
j++;
}
else modify_chars(chars,count,cur,i,j);
}
modify_chars(chars,count,cur,i,j);
return i;
}
void modify_chars(vector<char>& chars, int& count, char& cur,int& i, int& j)
{
if(count==1)
{
chars[i]=cur;
i++;
}
else
{
chars[i]=cur;
string s=to_string(count);
for(int j=0;j<s.size();j++)
{
i++;
chars[i]=s[j];
}
i++;
}
count=0;
cur=chars[j];
}
};