443. String Compression
作者:互联网
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.
Follow up:
Could you solve it using only O(1)
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".
Example 4:
Input: chars = ["a","a","a","b","b","a","a"] Output: Return 6, and the first 6 characters of the input array should be: ["a","3","b","2","a","2"]. Explanation: The groups are "aaa", "bb", and "aa". This compresses to "a3b2a2". Note that each group is independent even if two groups have the same character.
1 class Solution { 2 public int compress(char[] chars) { 3 if (chars == null || chars.length == 0) { 4 return 0; 5 } 6 7 int size = 0; 8 int currentCount = 1; 9 for (int i = 1; i < chars.length; i++) { 10 if (chars[i] != chars[i - 1]) { 11 size += 1; 12 if (currentCount != 1) { 13 size += length(currentCount); 14 currentCount = 1; 15 } 16 } else { 17 currentCount++; 18 } 19 } 20 size += 1; 21 if (currentCount != 1) { 22 size += length(currentCount); 23 } 24 return size; 25 } 26 27 private int length(int num) { 28 return String.valueOf(num).length(); 29 } 30 }
标签:currentCount,group,String,443,chars,character,length,array,Compression 来源: https://www.cnblogs.com/beiyeqingteng/p/14515393.html