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LeetCode-String Compression

程序员文章站 2024-03-12 23:34:33
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Description:
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.

题意:给定一个字符数组,将这个字符数组压缩,返回压缩后的长度;对应的压缩规则是,如果压缩的长度为1,则不进行压缩,否则压缩为压缩的字符后跟连续出现的次数;

解法:遍历每个字符,统计其连续出现的次数后保存在StringBuilder变量中,最后将其结果值依次付给字符数组前k个(k=StringBuilder变量的长度);

Java
class Solution {
    public int compress(char[] chars) {
        StringBuilder result = new StringBuilder();
        for (int i = 0; i < chars.length;) {
            int index = i;
            while (index < chars.length && chars[index] == chars[i]) index++;
            result.append(chars[i]);
            if (index - i > 1) {
                result.append(index - i);
            }
            i = index;
        }
        for (int i = 0; i < result.length(); i++) {
            chars[i] = result.charAt(i);
        }
        return result.length();
    }
}
相关标签: LeetCode Java