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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();
}
}