这不是家庭作业。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。
我在这里找不到类似的问题。
在一本初学者的书中，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。到处都有“魔术数字”，方法是“公共静态”等。我打算在算法工作后修复这些东西......
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);
}
}
在调试了几天后，每次调试会话代码都会变得更加复杂......任何帮助将不胜感激。提前致谢。