java如何将长度为20的数组中算出0-9数字个出现多少次？
最新回答
巷尾青苔
2025-04-28 11:09:03
public class ArrNumCount {
public static void main(String[] args) {
//无序的int[]
int[] arr = {18,5,6,4,4,1,1,2,2,0,1,2,3,4,5,6,7,8,9,10,11,12,13};
//从小到大排序
for(int i=0;i<arr.length-1;i++){
for (int j = 0; j < arr.length - 1 -i; j++) {
if(arr[j]>arr[j+1]) {
int temp=arr[j];
arr[j]=arr[j+1];
arr[j+1]=temp;
}
}
}
eveCount(arr);
sumCount(arr);
}
//每个数字出现的 个数
public static void eveCount(int arr[]) {
int[] count = {0,0,0,0,0,0,0,0};
for (int i = 0; i < arr.length; i++) {
if(arr[i]>0&&arr[i]<9) {
count[arr[i]-1] = ++count[arr[i]-1];
}
}
for (int i = 0; i < count.length; i++) {
System.out.println((i+1)+"出现的次数："+count[i]);
}
}
//0-9之间的数字总共有多少个
public static void sumCount(int arr[]) {
int count = 0;
for (int i = 0; i < arr.length; i++) {
if(arr[i]>0&&arr[i]<9) {
count++;
}
}
System.out.println("0-9之间的数字一共有："+count+"个");
}
}
深秋叶落黄
2025-04-28 12:45:00
刚写的，应该这个意思吧
蓝莓格格巫
2025-04-28 19:19:14
public static void main(String[] args) {
//无序的int[]
int[] arr = {18,5,6,4,4,1,1,2,2,0,1,2,3,4,5,6,7,8,9,10,11,12,13};
//从小到大排序
for(int i=0;i<arr.length-1;i++){
for (int j = 0; j < arr.length - 1 -i; j++) {
if(arr[j]>arr[j+1]) {
int temp=arr[j];
arr[j]=arr[j+1];
arr[j+1]=temp;
}
}
}
eveCount(arr);
sumCount(arr);
}
//每个数字出现的 个数
public static void eveCount(int arr[]) {
int[] count = {0,0,0,0,0,0,0,0};
for (int i = 0; i < arr.length; i++) {
if(arr[i]>0&&arr[i]<9) {
count[arr[i]-1] = ++count[arr[i]-1];
}
}
for (int i = 0; i < count.length; i++) {
System.out.println((i+1)+"出现的次数："+count[i]);
}
}
//0-9之间的数字总共有多少个
public static void sumCount(int arr[]) {
int count = 0;
for (int i = 0; i < arr.length; i++) {
if(arr[i]>0&&arr[i]<9) {
count++;
}
}
System.out.println("0-9之间的数字一共有："+count+"个");
}
}
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