LeetCode 443. String Compression

原题链接在这里：https://leetcode.com/problems/string-compression/

题目：

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.

题解：

Accumlate the count of repeating chars, if count > 1, append to the char.

To use constant space, first assign the count from the end by count % 10. Then reverse this part.

Time Complexity: O(chars.length). Space: O(1).

AC Java:

class Solution {
    public int compress(char[] chars) {
        if(chars == null || chars.length == 0){
            return 0;
        }
        
        int n = chars.length;
        int i = 0;
        int count = 0;
        int pos = 0;
        while(i < n){
            char cur = chars[i];
            while(i < n && chars[i] == cur){
                i++;
                count++;
            }
            
            chars[pos++] = cur;
            if(count > 1){
                int start = pos;
                while(count != 0){
                    chars[pos++] = (char)(count % 10 + '0');
                    count /= 10;
                }
                
                reverse(chars, start, pos - 1);
            }
            
            count = 0;
        }
        
        return pos;
    }
    
    private void reverse(char[] chars, int i, int j){
        while(i < j){
            swap(chars, i++, j--);
        }
    }
    
    private void swap(char[] chars, int i, int j){
        char temp = chars[i];
        chars[i] = chars[j];
        chars[j] = temp;
    }
}

类似Encode and Decode Strings, Count and Say, Design Compressed String Iterator.