题目描述:
找出字符串中第一个只出现一次的字符
Java实现:
import java.util.HashMap;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
while (sc.hasNext()) {
String str = sc.nextLine();
System.out.println(getFirstAppearOnceChar(str));
}
}
private static String getFirstAppearOnceChar(String str) {
char[] chars = str.toCharArray();
HashMap<Character, Integer> map = new HashMap<>();
for (int i = 0; i < chars.length; i++) {
if (map.containsKey(chars[i])) {
int count = map.get(chars[i])+1;
map.put(chars[i], count);
}else {
map.put(chars[i], 1);
}
}
for (int i = 0; i < chars.length; i++) {
if (map.get(chars[i]) == 1)
return chars[i]+"";
}
return "-1";
}
}
关键点:
- 遍历两次字符串,第一次构建每个字符和它出现次数的map,第二次找到第一个出现一次的字符