443. String Compression
Given an array of characters chars, compress it using the following algorithm:
Begin with an empty string s. For each group of consecutive repeating characters in chars:
- If the group’s length is 1, append the character to s.
- Otherwise, append the character followed by the group’s length.
The compressed string s should not be returned separately, but instead, be stored in the input character array chars. Note that group lengths that are 10 or longer will be split into multiple characters in chars.
After you are done modifying the input array, return the new length of the array.
You must write an algorithm that uses only constant extra space.
Example 1:
Input: chars = [“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: The groups are “aa”, “bb”, and “ccc”. This compresses to “a2b2c3”.
Example 2:
Input: chars = [“a”]
Output: Return 1, and the first character of the input array should be: [“a”]
Explanation: The only group is “a”, which remains uncompressed since it’s a single character.
Example 3:
Input: chars = [“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: The groups are “a” and “bbbbbbbbbbbb”. This compresses to “ab12”.
Constraints:
- 1 <= chars.length <= 2000
- chars[i] is a lowercase English letter, uppercase English letter, digit, or symbol.
From: LeetCode
Link: 443. String Compression
Solution:
Ideas：
- Iterate over each character in chars.
- For each character, count the number of times it repeats consecutively.
- Write the character and its count (if greater than 1) to the chars array at the current write position.
- Update the write position.
- After processing all characters, return the new length of the array.
Code:
int compress(char* chars, int charsSize) {
int writePos = 0, count = 1;
for (int i = 1; i <= charsSize; i++) {
// Check if the current character is the same as the previous one or if we've reached the end of the array
if (i < charsSize && chars[i] == chars[i - 1]) {
count++;
} else {
// Write the previous character and its count (if greater than 1) to the array
chars[writePos++] = chars[i - 1];
if (count > 1) {
// Convert count to string and write each digit to the array
char countStr[12]; // Enough to hold any integer
sprintf(countStr, "%d", count);
for (int j = 0; countStr[j] != '\0'; j++) {
chars[writePos++] = countStr[j];
}
}
count = 1; // Reset count for the next character
}
}
return writePos; // New length of the array
}
