我试图在最大值行中的列中获取最小值。我试图使用for循环,但它无法正常工作。
假设一个矩阵A:
3 4 5
2 3 4
1 2 3
我希望程序在[0]行中找到最大值,然后在max的colunm中找到最小值。 因此结果应该是:行[0] = 5的最大值,列[2]的最小值= 3。 然后我希望它对所有行做同样的事情,这就是为什么我会使用while循环。
这是矩阵:
public int[][] createMatrix(int a, int b){
Scanner inputm = new Scanner(System.in);
A = new int[a][b];
System.out.println("Enter elements for matrix A : ");
for (int i=0 ; i < A.length ; i++){
System.out.println("Enter numbers for " + i +"th row");
for (int j=0 ; j < A[i].length ; j++){
A[i][j] = inputm.nextInt();
}
}
return A;
}
public int[][] displayMatrix(){
System.out.println("Matrix A: ");
for (int i=0 ; i < A.length ; i++)
{ System.out.println();
for (int j=0 ; j < A[i].length ; j++){
System.out.print(A[i][j]+" ");
}
}
return A;
}
public int getMaximumOfEveryRow (int c){
a=c;
int i= 0;
int j;
while(i < A[a].length){
max = Integer.MIN_VALUE;
for ( j = 0; j < A [ i ].length; j++ )
if ( A [ i ] [ j ] > max ){
max = A [ i ] [ j ];
}
for ( i = 0; i < A [ i ].length; i++ )// e
if ( A [ i ] [ j ] < min ){
min = A [ i ] [ j ];
}
System.out.println( "\n Maximum of row " + j + " = " + max );
System.out.println( "Minimum of column " + i + " = " + min );
if(max == min){
System.out.println( min+ " = " + max );
System.out.println( "This is a saddle point. ");
}
i++;
}
return max;
}
这就是我到目前为止所做的:
public int getMaximumOfEveryRow (int c){
a=c;
int i= 0;
int j;
while(i < A[a].length){
max = Integer.MIN_VALUE;
for ( j = 0; j < A [ i ].length; j++ )
if ( A [ i ] [ j ] > max ){
max = A [ i ] [ j ];
}
int e = j;
int r;
for ( i = 0; i < A [ i ].length; i++ )// e
if ( A [ i ] [ j ] < min ){
min = A [ i ] [ j ];
}
System.out.println( "\n Maximum of row " + j + " = " + max );
System.out.println( "Minimum of column " + i + " = " + min );
if(max == min){
System.out.println( min+ " = " + max );
System.out.println( "This is a saddle point. ");
}
i++;
}
return max;
}
public int getMaximumOfEveryColumn ()
{
for ( int i = 0; i < A.length; i++ )
{
maxc = Integer.MIN_VALUE;
for ( int j = 0; j < A [ i ].length; j++ )
if ( A [ j ] [ i ] > maxc )
maxc = A [ j ] [ i ];
System.out.println( "Maximum of column " + i + " = " + maxc );
}
return maxc;
}
public int getMinimumOfEveryColumn_(){
for ( int i = 0; i < A.length; i++ )
{
minc = Integer.MAX_VALUE;
for ( int j = 0; j < A [ i ].length; j++ )
if ( A [ j ] [ i ] < minc )
minc = A [ j ] [ i ];
System.out.println( "Minimum of column " + i + " = " + minc );
}
return minc;
}
public int getMaximumOfEveryRow ()
{
for ( int i = 0; i < A.length; i++ )
{
maxr = Integer.MIN_VALUE;
for ( int j = 0; j < A [ i ].length; j++ )
if ( A [ i ] [ j ] > maxr )
maxr = A [ i ] [ j ];
System.out.println( "Maximum of row " + i + " = " + maxr );
}
return maxr;
}
用于在colunm中查找最大值,然后在该最大值的行中查找最小值的代码。
public void get_max_of_the_row_of_local_min ()
{
for ( int i = 0; i < A.length; i++ )
{
min = Integer.MAX_VALUE;
max = Integer.MIN_VALUE;
int index_of_min_in_its_col = 0;
//maxc = Integer.MIN_VALUE;
for ( int j = 0; j < A [ i ].length; j++ )
if ( A [ j ] [ i ] > max ){
max = A [ j ] [ i ];
index_of_min_in_its_col = i;
System.out.println( " Maximum of col [" + i + "] = " + max);
}
for ( int j = 0; j < A [ index_of_min_in_its_col ].length; j++ )
if ( A [ index_of_min_in_its_col ] [ index_of_min_in_its_col ] < min ){
min = A [ index_of_min_in_its_col ] [ index_of_min_in_its_col ];
a =j;
}
//System.out.print( " Maximum of col [" + j + "] = " + max);
System.out.println( " Minimum of rol [" + index_of_min_in_its_col + "] = " + min );
if(max == min){
System.out.println("This is a saddle point.");
}
}
答案 0 :(得分:1)
以下是解决方案:
public void get_minimum_of_the_column_of_local_maximum ()
{
for ( int i = 0; i < A.length; i++ )
{
min = Integer.MAX_VALUE;
max = Integer.MIN_VALUE;
int index_of_maximum_in_its_row = 0;
for ( int j = 0; j < A [ i ].length; j++ )
if ( A [ i ] [ j ] > max )
{
max = A [ i ] [ j ];
index_of_maximum_in_its_row = j;
}
for ( int j = 0; j < A [ index_of_maximum_in_its_row ].length; j++ )
if ( A [ j ] [ index_of_maximum_in_its_row ] < min )
min = A [ j ] [ index_of_maximum_in_its_row ];
System.out.print( " Maximum of row [" + i + "] = " + max);
System.out.println( " Minimum of column [" + index_of_maximum_in_its_row + "] = " + min );
}
}
此代码段的作用是什么?
您有min = Integer.MAX_VALUE, max = Integer.MIN_VALUE
个变量。我们将这些值分配给这个数字的原因是为了找到最大/最小值。怎么样?最初min
具有最大的整数值,但只要我们找到小于它的值,我们就会更新最小值,因此min
的值会减小。与max
变量相同的技术,但当然通过比较后增加值来实现另一个方向。
第一个内部for循环确定行 i 的最大值,然后使用变量index_of_maximum_in_its_row
标记该行中的最大索引。它需要稍后在第二个内循环中使用。
第二个内部循环确定列号index_of_maximum_in_its_row
的最小值,在该列上进行一次迭代。比该方法打印结果。
答案 1 :(得分:0)
这是最简单的实现。
for(int i=0; i<arr.length ; i++){
int columnNumber = getMaxElementsColumnNumber(arr,i); // i will represent row number
int minElement = getMinimumElementInColumn(arr,columnNumber);
}
getMaxElementsColumnNumber()将打印当前行中的最大数字,并将返回该元素的列号
getMinimumElementInColumn()将遍历矩阵的列并返回该列中的最小元素。
您可以在同一个for循环中实现代码。通过这种方式,您还可以检查鞍点。