LeetCode443. 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.

class Solution {
public:
	int compress(vector<char>& chars) {
		if (chars.size() <= 1) return chars.size();
		char pre = chars[0];
		int ans = 0;
		int count = 1;
		int idx = 0;
		for (int i = 1; i < chars.size();i++) {
			char c = chars[i];
			if (pre == c) count++;
			else {
				chars[idx++] = pre;
				if (count > 1) {
					string s = to_string(count);
					for (char tmp : s) chars[idx++] = tmp;
					ans += s.size()+1;
				}
				else ans += 1;
				pre = c;
				count = 1;
			}
		}
		chars[idx++] = pre;
		if (count > 1) {
			string s = to_string(count);
			for (char tmp : s) chars[idx++] = tmp;
			return ans + s.size()+1;
		}
		else return ans + 1;
	}
};