算法题一道:带层级的一维数组转换为森林
题目:将带层级的一维数组转换成森林(例如组织架构、品类、财务科目等)
class Node {
string id;
string parentId;
}
class TreeNode {
string id;
list<TreeNode> children;
}
//实现下方函数
public List<TreeNode> convertListToTree(List<Node> list) {
}
- 入参
[
{id:“s1”, parentId:“s2”},
{id:“s2”, parentId:“s3”},
{id:“s4”, parentId:“s3”},
{id:“s5”, parentId:“s6”},
{id:“s3”, parentId:null}
{id:“s6”, parentId:“s7”}
]
- 结果
[
{ id:“s3”, children:[
{id:“s2”, children:[
{id:“s1”}
]},
{id:“s4”, children:null},
]},
{ id:“s6”, children:[
{id:“s5”, children:null},
]},
]
- 代码实现:不缺点对不对,给的示例可以通过,请在评论区批评指正。
public class Main {
public static void main(String[] args) {
List<Node> list = new ArrayList<>();
list.add(new Node("s1","s2"));
list.add(new Node("s2","s3"));
list.add(new Node("s4","s3"));
list.add(new Node("s5","s6"));
list.add(new Node("s3","null"));
list.add(new Node("s6","s7"));
List<TreeNode> treeNodes = convertListToTree(list);
System.out.println();
}
//实现下方函数
public static List<TreeNode> convertListToTree(List<Node> list) {
HashMap<String, TreeNode> map = new HashMap<>();
List<TreeNode> treeNodeList = new ArrayList<>();
// 首先构造一个map
for (int i = 0; i < list.size(); i++) {
Node node = list.get(i);
String id = node.id;
String parentId = node.id;
TreeNode treeNode = new TreeNode();
treeNode.id = parentId;
treeNode.children = new ArrayList<>();
map.put(id, treeNode);
}
List<TreeNode> resultList = new ArrayList<>();
// 进行遍历,构造森林
for (int i = 0; i < list.size(); i++) {
Node node = list.get(i);
TreeNode curParentTreeNode = map.get(node.parentId);
// 如果取不到父节点,说明应该是根节点,本题存在多棵森林情况
if (curParentTreeNode == null){
TreeNode curTreeNode = map.get(node.id);
resultList.add(curTreeNode);
}else{
// 取到父节点,就将当前节点放到父节点的children集合中。
TreeNode curTreeNode = map.get(node.id);
List<TreeNode> children = curParentTreeNode.children;
children.add(curTreeNode);
}
}
return resultList;
}
}
class Node {
String id;
String parentId;
public Node(){
}
public Node(String id, String parentId){
this.id = id;
this.parentId = parentId;
}
}
class TreeNode {
String id;
List<TreeNode> children;
}