443. String Compression

443. String Compression (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:

1. All characters have an ASCII value in [35, 126].
2. 1 <= len(chars) <= 1000.

Companies:
Microsoft, Goldman Sachs, Bloomberg

Related Topics:
String

Similar Questions:

• Count and Say (Easy)
• Encode and Decode Strings (Medium)
• Design Compressed String Iterator (Easy)

Solution 1.

// OJ: https://leetcode.com/problems/string-compression/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(1)
class Solution {
public:
    int compress(vector<char>& chars) {
        int i = 0, N = chars.size(), j = 0, r = 0;
        while (i < N) {
            j = i + 1;
            while (j < N && chars[j] == chars[i]) ++j;
            chars[r++] = chars[i];
            if (j - i > 1) {
                int cnt = j - i, pow = 1;
                while (cnt / (pow * 10)) pow *= 10;
                while (pow) {
                    chars[r++] = '0' + cnt / pow;
                    cnt %= pow;
                    pow /= 10;
                }
            }
            i = j;
        }
        return r;
    }
};