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java - 查找矩阵中相邻数字的最大面积

转载 作者:塔克拉玛干 更新时间:2023-11-03 03:54:06 25 4
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这不是作业。我是编程初学者,这也是我在这里的第一篇文章 - 请多多包涵。

我无法在此处找到类似的问题。

在一本初学者的书上,我发现了以下问题:

# Find the biggest area of adjacent numbers in this matrix:
1 3 2 2 2 4
3 3 3 2 4 4
4 3 1 2 3 3 #--> 13 times '3'
4 3 1 3 3 1
4 3 3 3 1 1

这是我到目前为止的代码,使用来自 http://www.algolist.net/Algorithms/Graph_algorithms/Undirected/Depth-first_search 的 DFS 实现.到处都是“魔数(Magic Number)”,方法是“公共(public)静态”等。-我打算在算法工作后修复这些问题...

public class AdjacentAreaInMatrix {
/*
* Enums for the state of the Nodes, for use in DFS/BFS
*/
private enum NodeState {
Visited, InProgress, Unvisited
};

/*
* These 2 'magic' numbers come from the hardcoded 'matrix' below,
* cause it has 5 rows and 6 columns
*/
public static final int ROWSCOUNT = 5;
public static final int COLUMNSCOUNT = 6;

/*
* Two variables for counting the maximum sequence
* of numbers (as required by the problem definition)
*/
private static int tempElementsCount = 0;
private static int maxElementsCount = 1; // except if the matrix is empty, then it should be 0

/*
* The hardcoded matrix
*/
private static final int[][] matrix = new int[][] {
{ 1, 3, 2, 2, 2, 4 },
{ 3, 3, 3, 2, 4, 4 },
{ 4, 3, 1, 2, 3, 3 },
{ 4, 3, 1, 3, 3, 1 },
{ 4, 3, 3, 3, 1, 1 } };

/*
* Create an auxiliary matrix 'state' to implement DFS.
* Initialize the whole matrix as 'unvisited' and
* start DFS at the first element of the matrix
*/
public static void DFS() {
NodeState state[][] = new NodeState[ROWSCOUNT][COLUMNSCOUNT];
// clear the state of the matrix
for (int i = 0; i LT ROWSCOUNT; i++) {
for (int j = 0; j LT COLUMNSCOUNT; j++) {
state[i][j] = NodeState.Unvisited;
}
}
runDFS(0, 0, state);
}

/*
* Using the auxiliary matrix "state[][]", use DFS to traverse the
* 'real' matrix[][]
*/
public static void runDFS(int i, int j, NodeState state[][]) {
state[i][j] = NodeState.InProgress;
// traverse the whole matrix state[][] and recursively run runDFS() from the needed elements.
for (int rows = 0; rows LT ROWSCOUNT; rows++) {
for (int columns = 0; columns LT COLUMNSCOUNT; columns++) {
/*
* ----------------------------------------------------------------------
* For the logic in the 'if' statement regarding the adjacent elements:
* i0j0 i1j0 i1j0
* i0j1 i1j1 i2j1
* i0j2 i1j2 i2j2
* It uses the thing, that the sum of (i+j) for the coordinates of
* the elements above, below, on the left and on the right of i1j1
* are exactly +1/-1 of the sum of the coordinates of i1j1
* -> i1j2 to 1+2 = 3
* -> i2j1 to 1+2 = 3
* -> i1j1 to 1+1 = 2 (the current element) -> matrix[i][j]
* -> i1j0 to 1+0 = 1
* -> i0j1 to 1+0 = 1
* ----------------------------------------------------------------------
*/
if ((matrix[i][j] == matrix[rows][columns]) // if the values are equal
&& ((((i+j) - (rows + columns)) == 1) || (((i+j) - (rows + columns)) == -1))// and if the element is adjacent
&& (state[rows][columns] == NodeState.Unvisited)) { // and if the element is still not visited
tempElementsCount++;
if (tempElementsCount > maxElementsCount) {
maxElementsCount = tempElementsCount;
}
runDFS(rows, columns, state); // recursively run DFS for each element, that "isEdge"
} else {
// if the elements aren't [adjacent, equal and not visited], start the count again from '0'
tempElementsCount = 0;
}
}
}
state[i][j] = NodeState.Visited;
}

public static void go() {
AdjacentAreaInMatrix.DFS();
System.out.println(maxElementsCount);
}
}

在调试了几天之后,随着每次调试 session ,代码变得更加复杂...任何帮助将不胜感激。提前致谢。

最佳答案

我认为问题在于您每次都在重置 tempElementsCount。想象一下您的代码如何在给定的矩阵上工作,您会发现在 runDFS() 方法中您总是从元素 (0, 0) 开始搜索,对于该元素 if 子句将为假,因此您在之前重置 tempElementsCount您可以继续搜索其他(可能是相邻的)元素。希望我说得够清楚...

关于java - 查找矩阵中相邻数字的最大面积,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/1260901/

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