443. String Compression
Easy
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
.
package leetcode.easy; public class StringCompression { public int compress(char[] chars) { int indexAns = 0, index = 0; while (index < chars.length) { char currentChar = chars[index]; int count = 0; while (index < chars.length && chars[index] == currentChar) { index++; count++; } chars[indexAns] = currentChar; indexAns++; if (count != 1) { for (char c : String.valueOf(count).toCharArray()) { chars[indexAns] = c; indexAns++; } } } return indexAns; } @org.junit.Test public void test() { char[] chars1 = { ‘a‘, ‘a‘, ‘b‘, ‘b‘, ‘c‘, ‘c‘, ‘c‘ }; char[] chars2 = { ‘a‘ }; char[] chars3 = { ‘a‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘, ‘b‘ }; System.out.println(compress(chars1)); System.out.println(compress(chars2)); System.out.println(compress(chars3)); } }
原文地址:https://www.cnblogs.com/denggelin/p/11956851.html
时间: 2024-10-25 15:04:01