443.String-Compression

443. String Compression

题目地址

题目描述

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.

代码

Approach 1: Read and Write Heads

Intuition
We will use separate pointers read and write to mark where we are reading and writing from. Both operations will be done left to right alternately: we will read a contiguous group of characters, then write the compressed version to the array. At the end, the position of the write head will be the length of the answer that was written.
class Solution {
public int compress(char[] chars) {
int indexAns = 0, index = 0;
while (index < chars.length) {
char ch = chars[index];
int count = 0;
while (index < chars.length && chars[index] == ch) {
index++;
count++;
}
chars[indexAns++] = ch;
if (count != 1) {
for (char c : Integer.toString(count).toCharArray()) {
chars[indexAns++] = c;
}
}
}
}
}