HDU steps 1.2.7 AC Me

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分析：这个题算法上没有什么难度，主要在于考察基本字符串，基本控制的运用。在java中，1、有判断是否为字母的函数“Character.isLowerCase(s.charAt(i))” ；

2、map容器可以很方便的存储计算的结果，用到了map.put(),map.get(), map.containsKey(). 其中当想要修改value的值时，可以先get，再put覆盖。

3、get(Object key)
返回指定键所映射的值；如果此映射不包含该键的映射关系，则返回 null。

AC Me

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4644 Accepted Submission(s): 1620

Problem Description

Ignatius is doing his homework now. The teacher gives him some articles and asks him to tell how many times each letter appears.

It's really easy, isn't it? So come on and AC ME.

Input

Each article consists of just one line, and all the letters are in lowercase. You just have to count the number of each letter, so do not pay attention to other characters. The length of article is at most 100000. Process to the end of file.

Note: the problem has multi-cases, and you may use "while(gets(buf)){...}" to process to the end of file.

Output

For each article, you have to tell how many times each letter appears. The output format is like "X:N".

Output a blank line after each test case. More details in sample output.

Sample Input

hello, this is my first acm contest!work hard for hdu acm.

Sample Output

a:1b:0c:2d:0e:2f:1g:0h:2i:3j:0k:0l:2m:2n:1o:2p:0q:0r:1s:4t:4u:0v:0w:0x:0y:1z:0a:2b:0c:1d:2e:0f:1g:0h:2i:0j:0k:1l:0m:1n:0o:2p:0q:0r:3s:0t:0u:1v:0w:1x:0y:0z:0

Author

Ignatius.L

Source

杭州电子科技大学第三届程序设计大赛

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Ignatius.L

代码：

import java.util.HashMap;import java.util.Map;import java.util.Scanner;public class Main {public static void main(String[] args) {Map<Character, Integer> map = new HashMap<Character, Integer>();Scanner in = new Scanner(System.in);while (in.hasNext()) {map.clear();//System.out.println("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!");String s = in.nextLine();for (int i = 0; i < s.length(); i++) {if (Character.isLowerCase(s.charAt(i))) {if(map.containsKey(s.charAt(i))){int n = map.get(s.charAt(i));map.put(s.charAt(i), n+1);}else{map.put(s.charAt(i),1);}}}for(int i='a';i< 'a'+26;i++){if(map.containsKey((char)i)){System.out.println((char)i+":"+map.get((char)i));}else{System.out.println((char)i+":"+0);}}System.out.println();//System.out.println("----------------------------------");}}}

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