[leetcode] 443. String Compression

题目：

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.

代码：

class Solution {
2
public:
3
    int compress(vector<char>& chars) {
4
        if(chars.size() == 0) return 0;
5
        int cnt = 1;
6
        int place = 0;
7
        for(int i = 1; i < chars.size(); i++){
8
            if(chars[i] != chars[i-1]){
9
                chars[place++] = chars[i-1];
10
                if(cnt != 1){
11
                    string num = to_string(cnt);
12
                    for(int j = 0; j < num.size(); j++){
13
                        chars[place++] = num[j];
14
                    }
15
                    cnt = 1;
16
                }
17
            }else{
18
                cnt++;
19
            }
20
        }
21
        chars[place++] = chars[chars.size() - 1];
22
        if(cnt != 1){
23
            string num = to_string(cnt);
24
            for(int j = 0; j < num.size(); j++){
25
                chars[place++] = num[j];
26
            }
27
        }
28
        return place;
29
    }
30
};
  Custom