443.String Compression

443.String Compression

难度:Easy

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.

原地修改，需要注意重复元素出现次数，如果是1可以省略，后面补上的长度是次数的宽度。

class Solution {
public:
    int compress(vector<char>& chars) {
        int start=1;
        int count=1;
        int res=0;
        for(int i=1;i<chars.size();i++)
        {
            if(chars[i]==chars[i-1])
                count++;
            else
            {
                if(count==1){
                    res+=1;
                }
                else
                {
                    string tmp=to_string(count);
                    res+=1+tmp.length();
                    for(auto c:tmp)
                        chars[start++]=c;         
                    count=1;

                 }
                chars[start++]=chars[i];
            }                  
        }
        
         if(count==1){
                    res+=1;
                }
                else
                {
                    string tmp=to_string(count);
                    res+=1+tmp.length();
                    for(auto c:tmp)
                        chars[start++]=c;         
                 }
            return res;
    }
};

执行用时 :12 ms, 在所有 C++ 提交中击败了89.58%的用户 内存消耗 :9.1 MB, 在所有 C++ 提交中击败了91.06%的用户