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StringCompression.java
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StringCompression.java
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package string;
/**
* Created by gouthamvidyapradhan on 12/04/2018.
* 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.
Solution O(N) time complexity and O(1) space complexity. Maintain read and write pointers. Read from read pointer and
increment count when a repetition is found, when there is no repetition write the count value using write pointer.
*/
public class StringCompression {
/**
* Main method
* @param args
* @throws Exception
*/
public static void main(String[] args) throws Exception{
char[] A = {'a','a','b','b','c','c','c'};
System.out.println(new StringCompression().compress(A));
}
public int compress(char[] chars) {
int count = 0;
int i = 0;
int p = 0;
for(int j = 0; j < chars.length; j ++){
if(chars[i] == chars[j]){
count ++;
} else{
chars[p] = chars[i];
p++;
if(count > 1){
String countStr = String.valueOf(count);
for (int l = 0; l < countStr.length(); l++){
chars[p++] = countStr.charAt(l);
}
}
i = j;
count = 1;
}
}
chars[p] = chars[i];
p++;
if(count > 1){
String countStr = String.valueOf(count);
for (int l = 0; l < countStr.length(); l++){
chars[p++] = countStr.charAt(l);
}
}
return p;
}
}