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

Code:

class Solution {
    public int compress(char[] chars) {        
        int start = 0;
        for(int end = 0, count = 0; end < chars.length; end++) {
            count++;
            if(end == chars.length-1 || chars[end] != chars[end + 1] ) {
                //We have found a difference or we are at the end of array
                chars[start] = chars[end]; // Update the character at start pointer
                start++;
                if(count != 1) {
                    // Copy over the character count to the array
                    char[] arr = String.valueOf(count).toCharArray();
                    for(int i=0;i<arr.length;i++, start++)
                        chars[start] = arr[i];
                }
                // Reset the counter
                count = 0;
            }
        }
        return start;
    }
}

解题思路：

• 如果当前遍历元素不等于下一元素，则将此元素存入字符数组；
• 如果计数器count不等于1，则需要将count的每一位分开存入字符数组。