英文:
Count all amount of numbers in 2d array
问题
我正在尝试编写控制台扫雷游戏，但无法弄清如何迭代遍历二维数组，并计算每个元素周围的"-1"的数量。我的数组如下所示：
[0 0 0 0 0]
[0 0 -1 0 0]
[0 0 0 0 0]
[0 0 0 -1 -1]
[0 0 0 -1 -1]
其中"-1"表示地雷。我尝试使用以下代码简单地进行迭代：
for i := 0; i < row; i++ {
numb := 0
for j := 0; j < col; j++ {
if ary[i-1][j-1] == -1 {
numb++
}
}
if ary[i-1][j] == 1 {
numb++
}
//省略其余代码
}
但这导致了大量的代码，很难阅读和理解。有没有更灵活的方法，在"for-iterator"中检查current周围的所有元素，以便结果看起来像这样：
[0 1 1 1 0]
[0 1 -1 1 0]
[0 1 2 2 1]
[0 0 1 -1 -1]
[0 0 1 -1 -1]
英文:
I'm trying to write console minesweeper, but cant figure out how to iterate over 2d array, and count amount of "-1" around each of element. My array looks like
[0 0 0 0 0]
[0 0 -1 0 0]
[0 0 0 0 0]
[0 0 0 -1 -1]
[0 0 0 -1 -1]
Where "-1" that is mine. I've tried to simply iterate over it with code
for i := 0; i < row; i++ {
numb := 0
for j := 0; j < col; j++ {
if ary[i-1][j-1] == -1 {
numb ++ }
}
if ary[i-1][j] == 1 {
numb ++ }
//rest of code omitted
}
But that resulted in huge amout of code, which is hard to read and understand. Is there more flexible way, to check all elements around current in for-iterator, so in result it would look like
[0 1 1 1 0]
[0 1 -1 1 0]
[0 1 2 2 1]
[0 0 1 -1 -1]
[0 0 1 -1 -1]
答案1
得分: 2
for i := 0; i < 行数; i++ {
for j := 0; j < 列数; j++ {
if ary[i][j] == -1 {
for k := max(0, i-1); k <= min(行数-1, i+1); k++ {
for l := max(0, j-1); l <= min(列数-1, j+1); l++ {
if ary[k][l] != -1 {
ary[k][l]++
}
}
}
}
}
}
前两个外部循环遍历数组。如果观察到的数字是 -1，则增加周围的所有单元格，除非该单元格上有地雷。为了避免检查 ary[k][l] 是否超出边界，使用 min/max 的技巧在之前计算边界，以确保 k 和 l 不超出边界。例如，max(0, i-1) 将始终返回 i-1，除非 i-1 小于 0，此时返回 0。
英文:
for i := 0; i < row; i++ {
for j := 0; j < col; j++ {
if ary[i][j] == -1 {
for k := max(0, i - 1); k <= min(row - 1, i + 1); k++ {
for l := max(0, j - 1); l <= min(col - 1, j + 1); l++ {
if ary[k][l] != -1 {
ary[k][l] ++
}
}
}
}
}
}
The first two outer loops go through the array. If the observed number is -1 then it increases all surrounding cells unless there is a mine on that field. To avoid the checks if ary[k][l] is not outside of bounds the bounds are calculated before by using the trick with min / max to ensure that k and l are not outside of bounds. E.g. max(0, i-1) will always return i-1 unless i-1 is smaller than 0, then it returns 0.
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