以下是沙漏问题定义的链接: https://www.hackerrank.com/challenges/30-2d-arrays
我写了以下程序:
package day11;
import java.util.Scanner;
public class Solution {
public static void main(String ... args){
Scanner scan = new Scanner(System.in);
int[][] arr = new int[6][6];
int maxHourGlassValue = 0;
int temp = 0;
int currMax = 0;
int k = 0, l = 0;
for(int i = 0 ; i < 6 ; i++){
for(int j = 0 ; j < 6 ; j++){
arr[i][j] = scan.nextInt();
}
}
for(int i = 1 ; i < 5 ; i++){
for(int j = 1 ; j < 5 ; j++){
if(maxHourGlassValue < currMax){
maxHourGlassValue = currMax;
}
}
}
System.out.println(maxHourGlassValue);
}
}
我只能运行8个测试用例中的6个。什么可能出错????
答案 0 :(得分:1)
试试这段代码,我没有写那段代码,我只是复制它并从here修改它
Set basket = link.Execute("SELECT * FROM Table1;")
答案 1 :(得分:0)
以下是hourglass problem的所有测试用例成功运行的代码。
public static void main(String[] args) {
try (Scanner scan = new Scanner(System.in)) {
int[][] arr = new int[6][6];
int maxHourGlassValue = -63;//Assigning (-9*7=)-63 which is the minimum possible value of "hourglass sum".
//Reading inputs.
for (int i = 0; i < 6; i++) {
for (int j = 0; j < 6; j++) {
arr[i][j] = scan.nextInt();
}
}
//Logic.
/**
* Index of both i and j will run from 1 to 4 (one less than n-1 where n = 6)
* So for each i and j iteration calculating the sum of hourglass.
*/
int iHGValueTemp = 0;
for (int i = 1; i < 5; i++) {
for (int j = 1; j < 5; j++) {
iHGValueTemp = arr[i][j] + /*Main element*/
arr[i - 1][j - 1] + arr[i - 1][j] + arr[i - 1][j + 1]+ /*Top three elements of main element.*/
arr[i + 1][j - 1] + arr[i + 1][j] + arr[i + 1][j + 1]; /*Bottom three elements of main element.*/
if (iHGValueTemp > maxHourGlassValue) {
maxHourGlassValue = iHGValueTemp;
}
}
}
//Output.
System.out.println(maxHourGlassValue);
}
}
我只在评论中的代码中写了描述。如果有任何疑问,请参考并与我讨论。