Title

the problem link

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.

注意：是In-place

Solution

//cpp:
class Solution {
public:
	int compress(vector<char>& a) {
		int t = 0;
		int lasti=0;
		for (int i = 0; i < a.size(); i++) {
			if (i+1==a.size()||a[i+1] != a[i]) {
				a[t++] = a[i];
				if (i > lasti) {
					string len = to_string(i - lasti + 1);
					for (char c : len) a[t++] = c;
				}

				lasti = i + 1;
			}
		}
		for (auto it : a) {
			cout << it << " ";
		}
		cout << t;
		return t;
	}
};