对代码进行了一些编辑，试图找出为什么我的X＆＃39; s [-1]没有被包含在找到该行的平均值中。这是我的平均值。知道为什么不计算我的-1＆＃39;
输出[预期]：
USER INPUT: 3
O O O
X X X
X X X
TOTAL OPENNESS OF [I][J] = 1
TOTAL OPENNESS OF [I][J+1] = 2
TOTAL OPENNESS OF [I][J+2] = 1
TOTAL SUM AVERAGE FOR THAT ROW = 1.3
HOWEVER..FOR ROW 2 AND ROW 3
TOTAL SUM AVERAGE FOR THOSE ROWS = 0
WHICH IS INCORRECT IT SHOULD = -1
public static void openfactor(char[][] mazeValue, int n){
for(int i = 1; i<=n; i++)
{
double rowAvg=0;
double totalRowAvg=0;
for(int j=1;j<=n;j++)
{
int count=0;
int totalOpeness=0;
int totalRowOpeness = 0;
//double rowAvg=0;
if(mazeValue[i][j]=='X'){
System.out.println("tHIS IS AN X FOR : [" + i + "]" +"[" + j + "] IS -1 ");
count = -1;
}
else
{
//YOU NEED TO VERIFY THAT J IS NOT OUT OF BOUND
if( j-1>=1)
{
if(mazeValue[i][j-1]=='O')
count++;
}
// System.out.println("cout: "+count);
if(i-1>=1 && j-1>=1)
{
if(mazeValue[i-1][j-1]=='O')
count++;
}
// System.out.println("cout: "+count);
if(i-1>=1)
{
if(mazeValue[i-1][j]=='O')
count++;
}
// System.out.println("cout: "+count);
if(j+1<=n)
{
if(mazeValue[i][j+1]=='O')
count++;
}
// System.out.println("cout: "+count);
if(j+1<=n && i+1<=n)
{
if(mazeValue[i+1][j+1]=='O')
count++;
}
if (i+1<=n)
{
if(mazeValue[i+1][j]=='O')
count++;
}
// System.out.println("cout: "+count);
if(j-1>=1 && i+1<=n)
{
if(mazeValue[i+1][j-1]=='O')
count++;
}
if(i-1>=1 && j+1<=n)
{
if(mazeValue[i-1][j+1]=='O')
count++;
}
// System.out.println("cout: "+count);
totalOpeness = totalOpeness +count;
System.out.println("TOTAL OPENESS FOR : [" + i + "]" +"[" + j + "] IS " +totalOpeness);
totalRowOpeness = totalRowOpeness + totalOpeness;
//}//eND OF iF CONDITION\
}
rowAvg = (double)totalRowOpeness/(double)n;
System.out.println("ROW AVERAGE: "+rowAvg);
totalRowAvg = totalRowAvg + rowAvg;
System.out.println("SUM ROW AVERAGE: "+totalRowAvg);
}
System.out.println("TOTAL SUM ROW AVERAGE: " +totalRowAvg);
}
}
public static void printMaze(char mazeValue[][]) {
System.out.println("MAZE");
for (int i = 1; i < mazeValue.length; i++) {
for (int j = 1; j < mazeValue[i].length; j++) {
System.out.printf("%5c", mazeValue[i][j]);
}
System.out.printf("\n");
}
}
public static void main(String[] args) {
// TODO code application logic here
Scanner kbd = new Scanner(System.in);
System.out.println("ENTER A SINGLE INTEGER: ");
int n = kbd.nextInt();
char[][] mazeValue = new char[n + 1][n + 1];
System.out.println("ENTER A PATH: ");
for (int i = 0; i < mazeValue.length; i++) {
for (int j = 0; j < mazeValue[i].length; j++) {
if (i == 0 || j == 0 || i == n + 1 || j == n + 1)
mazeValue[i][j] = 'X';
else {
mazeValue[i][j] = kbd.next().charAt(0);
}
}
}
printMaze(mazeValue);
horizontalPath(mazeValue, n);
System.out.println(" ");
verticalPath(mazeValue,n);
System.out.println(" ");
openfactor(mazeValue, n);
}
}
答案 0 :(得分：0)
我不完全明白你想要完成什么，但我会假设你想要找到重复的值，使用下面的一些搜索算法做这个是二进制搜索的一个例子。希望它有所帮助。
import java.util.Scanner;
class BinarySearch
{
public static void main(String args[])
{
int c, first, last, middle, n, search, array[];
Scanner in = new Scanner(System.in);
System.out.println("Enter number of elements");
n = in.nextInt();
array = new int[n];
System.out.println("Enter " + n + " integers");
for (c = 0; c < n; c++)
array[c] = in.nextInt();
System.out.println("Enter value to find");
search = in.nextInt();
first = 0;
last = n - 1;
middle = (first + last)/2;
while( first <= last )
{
if ( array[middle] < search )
first = middle + 1;
else if ( array[middle] == search )
{
System.out.println(search + " found at location " + (middle + 1) + ".");
break;
}
else
last = middle - 1;
middle = (first + last)/2;
}
if ( first > last )
System.out.println(search + " is not present in the list.\n");
}
}
答案 1 :(得分：0)
这是您的请求的完整代码。你需要重新排序你的if语句你的逻辑是正确的： 这是输出：
MAZE
O O X
O O O
X X O
TOTAL OPENESS FOR : [0][0] IS 3
TOTAL OPENESS FOR : [0][1] IS 4
THERE IS AN X HERE FOR : [0][2]
Average of O's in this row is : 66.66667%
TOTAL OPENESS FOR : [1][0] IS 3
TOTAL OPENESS FOR : [1][1] IS 5
TOTAL OPENESS FOR : [1][2] IS 3
Average of O's in this row is : 100.0%
THERE IS AN X HERE FOR : [2][0]
THERE IS AN X HERE FOR : [2][1]
TOTAL OPENESS FOR : [2][2] IS 2
Average of O's in this row is : 33.333336%
这里是代码：
import java.util.Scanner;
public class sof {
public static boolean IsOutOfBound(int i, int j, int n)
{
if (i-1<1 || j-1<1 || i+1>n || j+1>n)
return true;
else
return false;
}
public static void openfactor(char[][] mazeValue, int n)
{
for(int i = 0; i<n; i++)
{
int TotalCounts=0;
for(int j=0;j<n;j++)
{
int count=0;
if(mazeValue[i][j]=='X'){
System.out.println("THERE IS AN X HERE FOR : [" + i + "]" +"[" + j + "] ");
//TotalCounts--;
}
else
{
//YOU NEED TO VERIFY THAT J IS NOT OUT OF BOUND
if( j-1>=0)
{
if(mazeValue[i][j-1]=='O')
count++;
}
// System.out.println("cout: "+count);
if(i-1>=0 && j-1>=0)
{
if(mazeValue[i-1][j-1]=='O')
count++;
}
// System.out.println("cout: "+count);
if(i-1>=0)
{
if(mazeValue[i-1][j]=='O')
count++;
}
// System.out.println("cout: "+count);
if(j+1<n)
{
if(mazeValue[i][j+1]=='O')
count++;
}
// System.out.println("cout: "+count);
if(j+1<n && i+1<n)
{
if(mazeValue[i+1][j+1]=='O')
count++;
}
if (i+1<n)
{
if(mazeValue[i+1][j]=='O')
count++;
}
// System.out.println("cout: "+count);
if(j-1>=0 && i+1<n)
{
if(mazeValue[i+1][j-1]=='O')
count++;
}
if(i-1>=0 && j+1<n)
{
if(mazeValue[j+1][i-1]=='O')
count++;
}
// System.out.println("cout: "+count);
//totalOpeness = totalOpeness +count;
System.out.println("TOTAL OPENESS FOR : [" + i + "]" +"[" + j + "] IS " + count);
TotalCounts++;
}//END OF else CONDITION
}//End of J loop
float Average = ((float)TotalCounts/(float)n) * 100;
System.out.println("Average of O's in this row is : " + Average+ "%");
}//End of I loop
}
public static void printMaze(char mazeValue[][],int n) {
System.out.println("MAZE");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.printf("%5c", mazeValue[i][j]);
}
System.out.printf("\n");
}
}
public static void main(String[] args) {
// TODO code application logic here
// TODO code application logic here
Scanner kbd = new Scanner(System.in);
System.out.println("ENTER A SINGLE INTEGER: ");
int n = kbd.nextInt();
char[][] mazeValue = new char[n][n];
System.out.println("ENTER A PATH: ");
for (int i = 0; i <n; i++) {
for (int j = 0; j < n; j++) {
//if (i == 0 || j == 0 || i == n + 1 || j == n + 1)
// mazeValue[i][j] = 'X';
// else {
mazeValue[i][j] = kbd.next().charAt(0);
// }
}
}
printMaze(mazeValue,n);
openfactor(mazeValue, n);
}
}