Description
Implement a method to perform basic string compression using the counts of repeated characters. For example, the string aabcccccaaa would become a2b1c5a3.
If the "compressed" string would not become smaller than the original string, your method should return the original string.
You can assume the string has only upper and lower case letters (a-z).
Example
str=aabcccccaaa return a2b1c5a3
str=aabbcc return aabbcc
str=aaaa return a4
解题：字符串压缩问题。用两个临时变量保存字符，一个变量count来计数，pre保存前一个字符，p保存当前字符，如果p不等于pre，将pre和count连到结果上去，并且跟新pre为p，一次循环即可。另外注意对最后一个（或几个相同的）字符进行处理。
public class Solution {
/**
* @param str: a string
* @return: a compressed string
*/
public String compress(String str) {
// write your code here
if(str == null)
return null;
String res = "";
int count = 0;
char pre = (byte)0;//前一个字符
char p = (byte)0;//当前字符
for(int i = 0; i < str.length(); i++){
if(pre == 0 && p == 0){
//初始化
pre = str.charAt(i);
p = str.charAt(i);
count++;
continue;//下一轮循环
}
//不是开头字符
p = str.charAt(i);
if(p == pre){
count++;
}else{
res = res + pre + String.valueOf(count);
pre = p;
count = 1;
}
}
//循环结束，处理最后一个（类）字符
res = res + pre + String.valueOf(count);
if(res.length() < str.length()){
return res;
}else{
return str;
}
}
}