Problem
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.
Solution
We can scan from left to right, remember two special positions: start of a repeating char sequence, end of the char sequence. The edge is detected with chars[i] != chars[start]. We also need to pay attention to the last position in the char array.
public class StringReduce {
public static int reduce(String str) {
char[] chars = str.toCharArray();
int anchor = 0;
int tail = 0;
//aaabbbccc
// .
for(int i = 0; i <= chars.length; i++) {//9
if(i == chars.length || chars[anchor] != chars[i]) {//6
chars[tail++] = chars[anchor];//a3b3c
if(i - anchor > 1) {
for(char c : (i - anchor +"").toCharArray()) {
chars[tail++] = c;//4
}
}
anchor = i;//6
}
}
for(int i = 0; i < tail; i++) {
System.out.print(chars[i]);
}
System.out.println();
return tail;
}
public static void main(String...args) {
System.out.println(reduce("aaabbbccc"));
System.out.println(reduce("abbbbbbbbbbbbbbbbbccc"));
System.out.println(reduce("bbbbbbbbbbbbbbbbb"));
}
}
public static int reduce(String str) {
char[] chars = str.toCharArray();
int anchor = 0;
int tail = 0;
//aaabbbccc
// .
for(int i = 0; i <= chars.length; i++) {//9
if(i == chars.length || chars[anchor] != chars[i]) {//6
chars[tail++] = chars[anchor];//a3b3c
if(i - anchor > 1) {
for(char c : (i - anchor +"").toCharArray()) {
chars[tail++] = c;//4
}
}
anchor = i;//6
}
}
for(int i = 0; i < tail; i++) {
System.out.print(chars[i]);
}
System.out.println();
return tail;
}
public static void main(String...args) {
System.out.println(reduce("aaabbbccc"));
System.out.println(reduce("abbbbbbbbbbbbbbbbbccc"));
System.out.println(reduce("bbbbbbbbbbbbbbbbb"));
}
}
The time cost is O(N), the space cost is O(1).
