443. String Compression

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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:

1
Could you solve it using only O(1) extra space?

Example 1:

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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:

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Input:
["a"]
Output:
Return 1, and the first 1 characters of the input array should be: ["a"]
Explanation:
Nothing is replaced.

Example 3:

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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.

解法1： in-place

inplace的解法就是用一个pos记录当前需要插入的位置。
用一个变量last记录上一个字符，用count记录那个字符重复的次数。
如果当前字符和last一样则更新count，否则的话就写入到array中。

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class Solution {
    public int compress(char[] chars) {
        if (chars == null || chars.length == 0) {
            return 0;
        }
        if (chars.length == 1) return 1;
        int pos = 0;
        char last = chars[0];
        int count = 1;
        for (int i = 1; i < chars.length; i++) {
            char current = chars[i];
            if (current == last) {
                count++;
            } else {
                chars[pos++] = last;
                if (count > 1) {
                    String digit = Integer.toString(count);
                    for (char d : digit.toCharArray()) {
                        chars[pos++] = d;
                    }
                }
                count = 1;
                last = current;
            }
        }
        chars[pos++] = chars[chars.length - 1];
        if (count > 1) {
            String digit = Integer.toString(count);
            for (char d : digit.toCharArray()) {
                chars[pos++] = d;
            }
        }
        return pos;
    }
}