LeetCode [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.

 

Constraints:

• 1 <= chars.length <= 2000
• chars[i].length == 1
• chars[i] is a lower-case English letter, upper-case English letter, digit or a symbol.

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class Solution { public int compress(char[] chars) { int n = chars.length; int ret = 0; int i = 0; while(i<n){ int j = i; while(j<n && chars[i]==chars[j]) j++; int count = j-i; chars[ret++] = chars[i]; if(count>1){ for(char c : Integer.toString(count).toCharArray()){ chars[ret++] = c; } } i = j; } return ret; } }