Java 循环和不变量?
我试图交换数组b的值,并在k中存储一个值,以便后条件b[0…h]9为真 到目前为止,我有:Java 循环和不变量?,java,loops,invariants,Java,Loops,Invariants,我试图交换数组b的值,并在k中存储一个值,以便后条件b[0…h]9为真 到目前为止,我有: for ( int b = 0; k!= b.length; k = k+1 ) { int p = k+1; for ( int h= k+1; h != b.length; h= h+1 ) { if ( b[h] < b[p] ){ p= h; } } int t = b[h]; b[h]= b[p
for ( int b = 0; k!= b.length; k = k+1 ) {
int p = k+1;
for ( int h= k+1; h != b.length; h= h+1 ) {
if ( b[h] < b[p] ){
p= h;
}
}
int t = b[h];
b[h]= b[p];
b[p]= t;
}
for(int b=0;k!=b.length;k=k+1){
int p=k+1;
对于(int h=k+1;h!=b.length;h=h+1){
如果(b[h]
据我所知,您试图做的是获取一个int
数组,对其排序,然后找到n>9
所在的索引
实现这一点最简单的方法是使用数组。sort
首先对数组进行排序,然后一个简单的for循环将显示最后一个索引的位置。然后可以将数组一分为二,使第一个数组包含所有值9
下面是一个例子:
public static void main(String[] args) {
//Test data
int[] arrayOfInt = {11, 42, 24, 1, 8, 9, 10, 15, 8, 29, 9, 34 };
//Sort the array
Arrays.sort(arrayOfInt);
int index = 0;
//This will get you the index where the last number is <= 9
for (int i = 0; i < arrayOfInt.length; i++) {
if (arrayOfInt[i] <= 9)
index = i + 1;
}
//Now you know that from arrayOfInt[0] -> arrayOfInt[index] will be <= 9
//If you want to use them seperately, you can just split the array into two
int[] lessThanOrEqualToNine = Arrays.copyOfRange(arrayOfInt, 0, index);
int[] greaterThanNine = Arrays.copyOfRange(arrayOfInt, index, arrayOfInt.length);
System.out.println(Arrays.toString(lessThanOrEqualToNine));
System.out.println(Arrays.toString(greaterThanNine));
/* Prints the output:
* [1, 8, 8, 9, 9]
* [10, 11, 15, 24, 29, 34, 42]
*/
}
publicstaticvoidmain(字符串[]args){
//测试数据
int[]arrayOfInt={11,42,24,1,8,9,10,15,8,29,9,34};
//对数组进行排序
Arrays.sort(arrayOfInt);
int指数=0;
//这将为您提供索引,其中最后一个数字是我已经编辑了我的代码,但我不确定在哪里包含9。您的代码无法编译,您的问题很模糊。请参阅