深夜水题,秒一波!!!
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
/**
* 题意:去掉字符串中重复的字符,如果去重之后的字符串长度为偶数,则输出“CHAT WITH HER!”,否则输出“IGNORE HIM!”
*
* @author TinyDolphin
* 2017/6/24 22:45.
*/
public class Main {
/**
* 字符串去除重复字符的方法
*/
public static String distinct(String str){
String distinctLetters; //去重之后的字符串
distinctLetters = str.charAt(0) + "";
char temp;
for (int index = 0; index < str.length(); index++) {
temp = str.charAt(index);
if (!distinctLetters.contains(temp + "")) {
distinctLetters += temp;
}
}
return distinctLetters;
}
public static void main(String[] args) throws IOException {
StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
PrintWriter out = new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
String letters; //输入的字母
while (in.nextToken() != StreamTokenizer.TT_EOF) {
letters = in.sval;
out.println(distinct(letters).length()%2==0?"CHAT WITH HER!":"IGNORE HIM!");
}
out.flush();
}
}
本文介绍了一种通过Java实现的字符串去重算法,并基于去重后的字符串长度奇偶性输出不同结果。具体而言,该算法首先遍历字符串,然后利用StringBuilder来记录不重复的字符,最终比较去重后字符串长度的奇偶性。
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