crack the coding interview 数组与字符串 1.5

crack the coding interview 数组与字符串 1.5

题目：
压缩字符串

package ac.crack_code_interview.ch01_array_string;import org.junit.Test;/** * description: * * @author liyazhou * @since 2017-06-27 10:45 * 9.1 数组与字符串 * * 题目 1.5 *      利用字符串重复出现的次数，编写一个方法，实现基本的字符串压缩功能。 *      比如，字符串 aabcccccaaa 会变成 a2b1c5a3。 *      若“压缩”后的字符串没有变短，则返回原先的字符串。 * * 思路 *      1. 先遍历字符串，计算压缩后的空间大小 * */public class Test05 {    // 是否可用 StringBuilder 类    public String compress2(String str){        if (str == null) return null;        StringBuilder sBuilder = new StringBuilder();        char currChar = str.charAt(0);        int counter = 1;        for (int i = 1; i < str.length(); i ++){            if (str.charAt(i) == currChar) counter ++;            else {                sBuilder.append(currChar).append(counter);                currChar = str.charAt(i);                counter = 1;            }        }        sBuilder.append(currChar).append(counter);        return sBuilder.length() < str.length() ? sBuilder.toString() : str;    }    public char[] compress(char[] chars){        if (chars == null) return chars;        int newLen = calculateLen(chars);        if (newLen > chars.length) return chars;        char[] compressedChars = new char[newLen];        int newIdx = 0;        char currChar = chars[0];        int counter = 1;        for (int i = 1; i < chars.length; i ++){            if (chars[i] == currChar) counter ++;            else {                compressedChars[newIdx++] = currChar;                String str = String.valueOf(counter);                for (int j = 0; j < str.length(); j ++)                    compressedChars[newIdx++] = str.charAt(j);                currChar = chars[i];                counter = 1;            }        }        compressedChars[newIdx++] = currChar;        String str = String.valueOf(counter);        for (int j = 0; j < str.length(); j ++)            compressedChars[newIdx++] = str.charAt(j);        return compressedChars;    }    private int calculateLen(char[] chars) {        int total = 0;        char currChar = chars[0];        int counter = 1;        for (int i = 1; i < chars.length; i ++){            if (chars[i] == currChar) counter ++;            else{                total ++;  // 当前字符占一个位置                total += String.valueOf(counter).length(); // 当前字符重复的次数是几位数字，则占几位                currChar = chars[i];                counter = 1;            }        }        total ++;        total += String.valueOf(counter).length();        return total;    }    @Test    public void test(){        String[] strs = {                "aaabbbcccddd",                "abcdefghijklmn"        };        for (String str : strs){            System.out.println(str);            char[] result = compress(str.toCharArray());            System.out.println(new String(result));            System.out.println(compress2(str));            System.out.println();        }    }}