关于java se 的小问题

Question

java语言程序设计第7章程序清单7-4

package Program7;

import Program6.Program6_10GetCharacter;

public class Program7_4 {

    public static void main(String[] args) {
        char[] chars = createArray();

        System.out.println("The lowercase letters are: ");
        displayArray(chars);

        int[] counts = countLetters(chars);

        System.out.println();
        System.out.println("The occurrences of each letters are: ");
        displayCounts(counts);

    }

    public static char[] createArray() {
        char[] chars = new char[100];

        for (int i = 0; i < chars.length; i++)
            chars[i] = Program6_10GetCharacter.getRandomLowerCaseLetter();

        return chars;
    }

    public static void displayArray(char[] chars) {
        for (int i = 0; i < chars.length; i++) {
            if((i + 1) % 20 == 0)
                System.out.println(chars[i]);
            else
                System.out.print(chars[i] + " ");
        }
    }

    public static int[] countLetters(char[] chars) {
        int[] counts= new int[26];

        for (int i = 0; i < chars.length; i++) {
            counts[chars[i] - 'a']++;
        }
        return counts;
    }

    public static void displayCounts(int[] counts) {
        for (int i = 0; i < counts.length; i++) {
            if((i + 1) % 10 == 0)
                System.out.println(counts[i] + " " + (char)(i + 'a'));
            else
                System.out.print(counts[i] + " " + (char)(i + 'a') + " ");              
        }

    }
}

在 public static int[] countLetters(char[] chars) 方法里，用for (int i = 0; i < chars.length; i++) {counts[chars[i] - 'a']++;}可以实现对字符的计数，没能理解。哪位大神能说明一下么。书上解释没看懂。

Answer 1

int[] counts= new int[26];counts这个数组有26个元素，的0下标存储的是a的个数，1下标存储b的个数，2下标存储c的个数……
chars[i] - 'a'，把这个字符转换成对应的下标，比如chars[i]是'a'，那么'a'-'a'=0，如果chars[i]是'b'，那么'b'-'a'=1，等等。
所以让对应下标++实现了对这个字母的计数。