我有一个表示为Set<Integer>[]
以下Set<Integer>[]：
[ { 1 }, { 2, 3 }, { 4 }, { 5, 6, 7 } ]
代表以下树：
1
/ \
/ \
/ \
2 3
| |
4 4
/|\ /|\
5 6 7 5 6 7
因此，树中的每个级别都被编码为Set。树中特定级别的所有子集都是相同的。第一组中可以有多个整数。
我想从Set<Integer>[]获取从根到离开的所有路径的列表：
[ [ 1, 2, 4, 5 ], [ 1, 2, 4, 6 ], [ 1, 2, 4, 7 ], [ 1, 3, 4, 5 ], [ 1, 3, 4, 6 ], [ 1, 3, 4, 7 ] ]
答案 0 :(得分：4)
在树中搜索的密钥通常是实现良好的Ajacency函数，该函数为特定节点提供相邻节点。
对于此树，邻接函数需要查找节点所在的级别，然后将下一级别的节点作为邻接返回。它看起来像这样：
/**
* This returns the adjacent nodes to an integer node in the tree
* @param node
* @return a set of adjacent nodes, and null otherwise
*/
public Set<Integer> getAdjacentsToNode(Integer node) {
for (int i = 0; i < tree.size(); i++) {
Set<Integer> level = tree.get(i);
if (level.contains(node) && i < tree.size() - 1) {
return tree.get(i + 1);
}
}
return null;
}
假设我们已经将树定义为类中的字段。
在我们运行搜索之前，我们希望适当地设置根目录并对路径做一些预备工作。因此，我们执行以下操作：
/**
* initializes our search, sets the root and adds it to the path
*/
public void initialize() {
root = -1;
for (Integer node : tree.get(0)) {
root = node;
}
currentPath.add(root);
}
假设currentPath和root已经定义为字段。
然后，当我们遍历树时，我们在树上运行DFS搜索，将每个节点附加到当前路径，并将该路径添加到我们的设置路径，并在到达deadend（叶子）时重置它：
/**
* runs a DFS on the tree to retrieve all paths
* @param tree
*/
public void runDFS(Integer node) {
if (getAdjacentsToNode(node) != null) {
for (Integer adjNode : getAdjacentsToNode(node)) {
currentPath.add(adjNode);
runDFS(adjNode);
}
currentPath.remove(currentPath.size() -1);
} else {
paths.add(currentPath.toArray(new Integer[0]));
currentPath.remove(currentPath.size() -1);
}
}
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashSet;
import java.util.List;
import java.util.Set;
public class DFS {
private List<Integer> currentPath = new ArrayList<Integer>();
private List<Integer[]> paths = new ArrayList<Integer[]>();
private ArrayList<Set<Integer>> tree;
private Integer root;
/**
* constructor
* @param tree
*/
public DFS(ArrayList<Set<Integer>> tree) {
this.tree = tree;
}
public List<Integer[]> getPaths() {
return paths;
}
public void setPaths(List<Integer[]> paths) {
this.paths = paths;
}
public Integer getRoot() {
return root;
}
public void setRoot(Integer root) {
this.root = root;
}
/**
* initializes our search, sets the root and adds it to the path
*/
public void initialize() {
root = -1;
for (Integer node : tree.get(0)) {
root = node;
}
currentPath.add(root);
}
/**
* This returns the adjacent nodes to an integer node in the tree
* @param node
* @return a set of adjacent nodes, and null otherwise
*/
public Set<Integer> getAdjacentsToNode(Integer node) {
for (int i = 0; i < tree.size(); i++) {
Set<Integer> level = tree.get(i);
if (level.contains(node) && i < tree.size() - 1) {
return tree.get(i + 1);
}
}
return null;
}
/**
* runs a DFS on the tree to retrieve all paths
* @param tree
*/
public void runDFS(Integer node) {
if (getAdjacentsToNode(node) != null) {
for (Integer adjNode : getAdjacentsToNode(node)) {
currentPath.add(adjNode);
runDFS(adjNode);
}
currentPath.remove(currentPath.size() -1);
} else {
paths.add(currentPath.toArray(new Integer[0]));
currentPath.remove(currentPath.size() -1);
}
}
}
为了测试它，我们试试这个：
public static void main(String[] args) {
ArrayList<Set<Integer>> tree = new ArrayList<Set<Integer>>();
Set<Integer> level1 = new HashSet<Integer>();
level1.add(new Integer(1));
Set<Integer> level2 = new HashSet<Integer>();
level2.add(new Integer(2));
level2.add(new Integer(3));
Set<Integer> level3 = new HashSet<Integer>();
level3.add(new Integer(4));
Set<Integer> level4 = new HashSet<Integer>();
level4.add(new Integer(5));
level4.add(new Integer(6));
level4.add(new Integer(7));
tree.add(level1);
tree.add(level2);
tree.add(level3);
tree.add(level4);
DFS dfsSearch = new DFS(tree);
dfsSearch.initialize();
dfsSearch.runDFS(dfsSearch.getRoot());
for (Integer[] path : dfsSearch.getPaths()) {
System.out.println(Arrays.toString(path));
}
我们得到以下输出：
[1, 2, 4, 5]
[1, 2, 4, 6]
[1, 2, 4, 7]
[1, 3, 4, 5]
[1, 3, 4, 6]
[1, 3, 4, 7]
答案 1 :(得分：0)
我不打算编写代码，但最简单的方法是深度优先遍历，在每个级别将每个条目附加到当前路径。
此外，您的返回值将是列表的集合（或数组）（因为垂直路径不能是无序集合）。
伪代码中的：
def getPaths(levels) {
return getPaths(levels, 0, new Set())
}
def getPaths(levels, currentIndex, paths) {
if(currentIndex == levels.length)
return paths
def newPaths = new Set(paths)
for(path : paths) {
for(level : levels) {
newPaths.add( path + level )
}
}
return getPaths(levels, currentIndex + 1, newPaths)
}
答案 2 :(得分：0)
尝试这样的事情：
public static List<Integer[]> getAllPaths(Set<Integer>[] tree){
// Get the overall number of path
int totalSize = 1;
for(Set<Integer> line : tree){
totalSize *= line.size();
}
// Create the empty paths
List<Integer[]> allPaths = new ArrayList<Integer[]>(totalSize);
for(int i = 0 ; i<totalSize ; ++i){
Integer[] path = new Integer[tree.length];
allPaths.add(path);
}
// Fill the paths
int indexLine = 0;
for (Set<Integer> line : tree) {
Iterator<Integer[]> pathIterator = allPaths.iterator();
Iterator<Integer> lineIterator = line.iterator();
while(pathIterator.hasNext()){
if(!lineIterator.hasNext()){
lineIterator = line.iterator();
}
pathIterator.next()[indexLine] = lineIterator.next();
}
++indexLine;
}
return allPaths;
}
它适用于您的示例：
public static void main(String[] args) {
Set<Integer> line1 = new HashSet<Integer>();
line1.add(new Integer(1));
Set<Integer> line2 = new HashSet<Integer>();
line2.add(new Integer(2));
line2.add(new Integer(3));
Set<Integer> line3 = new HashSet<Integer>();
line3.add(new Integer(4));
Set<Integer> line4 = new HashSet<Integer>();
line4.add(new Integer(5));
line4.add(new Integer(6));
line4.add(new Integer(7));
Set[] test = {line1, line2,line3,line4};
List<Integer[]> allPaths = getAllPaths(test);
for(Integer[] path : allPaths){
System.out.println(Arrays.toString(path));
}
}