Question

输入："Geeksforgeeks"。
输出应为G:2 e:4 k:2 s:2 f:1 o:1 r:1

public static void main(String[] args) 
{
    String str = "Geeksforgeeks";     ``
    char[] ch = new char[str.length()];
    int k=0;
    int[] f = new int[str.length()]; 
    for(int i = 0;i<str.length();i++){
        int add= 0;
        ch[k] = str.charAt(i);
        int flag =-1;
        for(int p = 0;p<k;p++)
        {
            if(ch[p].equalsIgnoreCase(str.charAt(i)))
            {
                flag++;
                break;
            }
        }
        if(flag == -1)
        {
            for(int j=i;j<str.length();j++)
            {
                if(ch[k].equalsIgnoreCase(str.charAt(j)))
                {
                    add++;
                }
            }
            f[k] = add;
            k++;
        }
    }
    for(int i = 0;i<k;i++)
    {
        System.out.println(ch[i]+ "    "+f[i]);
    }
}

Answer 1

char是原始类型，不能在其上使用equalsIgnoreCase。

要解决此问题，您可以将其解析为字符串：

String.valueOf(ch[p]).equalsIgnoreCase(str.charAt(i));

Answer 2

您可以像这样编写您的课程：

public class TestClass  {

static final int MAX_CHAR = 256;

static void getOccuringChar(String str)
{
    // Create an array of size 256 i.e. ASCII_SIZE
    int count[] = new int[MAX_CHAR];

    int len = str.length();

    // Initialize count array index
    for (int i = 0; i < len; i++)
        count[str.charAt(i)]++;

    // Create an array of given String size
    char ch[] = new char[str.length()];
    for (int i = 0; i < len; i++) {
        ch[i] = str.charAt(i);
        int find = 0;
        for (int j = 0; j <= i; j++) {

            // If any matches found
            if (str.charAt(i) == ch[j]) 
                find++;                
        }

        if (find == 1) 
            System.out.println("Number of Occurrence of " +
             str.charAt(i) + " is:" + count[str.charAt(i)]);            
       }        
      }
     }

和主要方法中：

    public static void main(String[] args){

    String str = "geeksforgeeks";
    TestClass.getOccuringChar(str);

    }

结果将是这样的：

Number of Occurrence of g is:2

Number of Occurrence of e is:4

Number of Occurrence of k is:2

Number of Occurrence of s is:2

Number of Occurrence of f is:1

Number of Occurrence of o is:1

Number of Occurrence of r is:1

错误“无法取消引用字符”

2 个答案: