Ok, now it seems you have come with an idea up, and it is basically a working idea.
It has just one flaw: The neigborhood your are looking at is wrong: for a point (i,j) the (von Neumann) neighbors are (i+1,j), (i,j+1), (i-1, j), and (i,j-1).
So when you check for this point it should in general work, BUT: You have to take special care of the borders. As you have no -1st and no n+1th column/row you have when you are there take that neighbor point out of consideration. How you handle it, depends on how you want to treat it: Should the neighborhood wrap around, should the border have a constant value, should it be treated as -infty?
This assumes you want to check the element's 8 immediate neighbours, i.e. straight and diagonal. Adjust if this is not what you want.
Also the requirement is finding the local maxima. This identifies a local minimum. Change < to >.
Another thing is that locals.add(i + j) doesn't do what you think it does. If element (i=3,j=4) is a local maximum, then you're saying locals.add(7), which clearly is not what you want.
I think that there is no need for 8 comparisons for every matrix element. You should take two extra arrays:
MATRIX[n+2][n+2]: It will contain yout input matrix with an extra layer of all INT_MIN's(So, that every matrix element has all 8 neighbours).
Mark[n+2][n+2] = {0} (You should only have to visit the elements in the input matrix which are not marked) If a matrix element is declared as a local minimum, you should marks all its neighbour in the Mark[][] vector.
By doing so, complexity will remain O(n^2) but no. of comparisons will be minimised.
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好吧，现在看来你已经有了一个想法，而且它基本上是一个可行的想法。
它只有一个缺陷：您正在查看的邻域是错误的：对于点 (i,j)，（冯诺依曼）邻居是 (i+1,j)、(i,j+1)、(i-1) ，j）和（i，j-1）。
因此，当您检查这一点时，它通常应该有效，但是：您必须特别注意边界。由于您没有 -1 列/行，也没有第 n+1 列/行，所以当您在那里时，请不考虑该邻居点。你如何处理它，取决于你想如何对待它：邻域是否应该环绕，边界是否应该具有恒定值，是否应该被视为-infty？
Ok, now it seems you have come with an idea up, and it is basically a working idea.
It has just one flaw: The neigborhood your are looking at is wrong: for a point (i,j) the (von Neumann) neighbors are (i+1,j), (i,j+1), (i-1, j), and (i,j-1).
So when you check for this point it should in general work, BUT: You have to take special care of the borders. As you have no -1st and no n+1th column/row you have when you are there take that neighbor point out of consideration. How you handle it, depends on how you want to treat it: Should the neighborhood wrap around, should the border have a constant value, should it be treated as -infty?
您可以编写
检查该元素是否小于其三个邻居的元素。那么其他五个邻居呢？让我们也检查一下它们：
这假设您想要检查元素的 8 个直接邻居，即直线和对角线。如果这不是您想要的，请进行调整。
还要求找到局部最大值。这确定了局部最小值。将
<更改为>。另一件事是
locals.add(i + j)并没有像你想象的那样做。如果元素 (i=3,j=4) 是局部最大值，那么您所说的是locals.add(7)，这显然不是您想要的。You write
which checks for whether the element is smaller than three of its neighbours. What about the other five neighbours? Let's check them, too:
This assumes you want to check the element's 8 immediate neighbours, i.e. straight and diagonal. Adjust if this is not what you want.
Also the requirement is finding the local maxima. This identifies a local minimum. Change
<to>.Another thing is that
locals.add(i + j)doesn't do what you think it does. If element (i=3,j=4) is a local maximum, then you're sayinglocals.add(7), which clearly is not what you want.MATRIX[n+2][n+2]：它将包含您的输入矩阵以及一个额外的数组所有 INT_MIN 的层（因此，每个矩阵元素都具有所有 8
邻居）。
Mark[n+2][n+2] = {0}（您只需访问未标记的输入矩阵）如果矩阵元素是
声明为局部最小值，您应该将其所有邻居标记为
Mark[][] 向量。
MATRIX[n+2][n+2]: It will contain yout input matrix with an extralayer of all INT_MIN's(So, that every matrix element has all 8
neighbours).
Mark[n+2][n+2] = {0}(You should only have to visit the elements inthe input matrix which are not marked) If a matrix element is
declared as a local minimum, you should marks all its neighbour in
the Mark[][] vector.
您必须分别存储 i 和 j 才能正确记住元素的位置。因此，
不会起作用。
你可以用它
来代替。要添加新发现的局部最大值的位置，请使用
You'll have to store i and j separately to remember the position of an element correctly. Thus,
won't work.
You could use
instead. And to add the position of a newly found local maximum, use