語言-關於java se 的小問題

關於java se 的小問題

java語言程序設計第7章程序清單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) 方法裡，用for (int i = 0; i < chars.length; i++) {counts[chars[i] - 'a']++;}可以實現對字符的計數，沒能理解。哪位大神能說明一下麼。書上解釋沒看懂。

最佳回答：

int[] counts= new int[26];counts這個數組有26個元素，的0下標存儲的是a的個數，1下標存儲b的個數，2下標存儲c的個數……
chars[i] - 'a'，把這個字符轉換成對應的下標，比如chars[i]是'a'，那麼'a'-'a'=0，如果chars[i]是'b'，那麼'b'-'a'=1，等等。
所以讓對應下標++實現了對這個字母的計數。