试图获得数组中的最大/最小值

时间:2012-02-24 02:33:36

标签: java arrays matrix max min

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.");
    }

 }

2 个答案:

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。