- Java code
String x ="_hjk2io_jie@n3hf4_"; //一个字符串。
把下划线转换为@，把@转换为下划线，最后是下划线结尾的话就去掉它，出现数字的时候表示的意思是数字后面的字母打印三次。
转换后的结果是：
- Java code
x="@hjkiio@jie_nhhhf4"
好像不是很简单。
------解决方案--------------------
1、难吗？
2、按题目的要求，你的答案应该是@hjkiiio@jie_nhhhf4
- Java code
String x = "_hjk2io_jie@n3hf4_"; if (x.charAt(x.length() - 1) == '_') x = x.substring(0, x.length() - 1); char[] temp = x.toCharArray(); StringBuffer y = new StringBuffer(); for (int i = 0; i < x.length(); i++) { if (temp[i] == '_') y.append('@'); else if (temp[i] == '@') y.append('_'); else if (temp[i] >= '0' && temp[i] <= '9' && i < x.length() - 1 && ('0' < temp[i + 1] || temp[i + 1] < '9')) y.append(temp[i + 1]).append(temp[i + 1]); else y.append(temp[i]); } System.out.println(y);
------解决方案--------------------
- Java code
public class Test{ public static String change(String str) { StringBuilder strBuilder = new StringBuilder(str); if (strBuilder.charAt(strBuilder.length() - 1) == '_') { strBuilder.setLength(strBuilder.length() - 1); } for (int i = 0; i < strBuilder.length(); i++) { if (strBuilder.charAt(i) == '_') { strBuilder.setCharAt(i, '@'); } else if (strBuilder.charAt(i) == '@') { strBuilder.setCharAt(i, '_'); } else if (Character.isDigit(strBuilder.charAt(i)) && i != strBuilder.length() - 1) { int t = Integer.parseInt(strBuilder.substring(i, i + 1)); strBuilder.deleteCharAt(i); for (int j = 1; j < t; j++) { strBuilder.insert(i, strBuilder.charAt(i)); } } } return strBuilder.toString(); } public static void main(String[] args) { String str = "_hjk2io_jie@n3hf4_"; System.out.println(change(str)); }}
------解决方案--------------------
修正一下，当然用上正则能清晰很多，但是我看见正则就头晕!
- Java code
String x = "_hjk2io_jie@n3hf4_"; if (x.charAt(x.length() - 1) == '_') x = x.substring(0, x.length() - 1); char[] temp = x.toCharArray(); StringBuffer y = new StringBuffer(); for (int i = 0; i < x.length(); i++) { if (temp[i] == '_') y.append('@'); else if (temp[i] == '@') y.append('_'); else if ((temp[i] >= '0' && temp[i] <= '9') && i < x.length() - 1 && (('a' < temp[i + 1] || temp[i + 1] < 'z') && ('A' < temp[i + 1] || temp[i + 1] < 'Z'))) y.append(temp[i + 1]).append(temp[i + 1]); else y.append(temp[i]); } System.out.println(y);