java - 层次结构数据转移

Question

在运行递归函数以获得员工/经理家谱后 - 进一步要求保留一个整体经理结构。

所以我会想象输入数组看起来像这样

[["Employee A", "1000", "Employee B", "1001", "Employee C", "1002"],
["Employee D", "1003", "Employee C", "1002"]]

并且输出数组需要看起来像这样

[["Employee A", "1000", "Employee B", "1001", "Employee C", "1002"],
["Employee D", "1003", null, null, "Employee C", "1002"]]

层级需要按照这种方式排序，以表明员工C始终保留高级经理的角色

public void refactorArray(String jsonData) throws Exception {
    JSONArray array = new JSONArray(jsonData);
    for (int i = 0; i < array.length(); i++) {

        //flag previous position of grandfather manager and name/id

        // if previous position does not match current position - do logic to pop the array at that element to put it back into the previous position
    }
}

Answer 1

基于首先找到最长的一个的想法，并将其作为 future boss-padding 的引用，产生以下代码，它适用于上述情况。

我们的想法是建立一个“老板”查找表，从我们找到的最长的叶到根路径构建。每当查找表中有老板时，我们都会确保他出现在最长路径中出现的相同位置，必要时用空值填充。

import org.json.JSONArray;
import java.util.ArrayList;
import java.util.HashMap;

public class T {
    static String refactor(String jsonData) {
        JSONArray array = new JSONArray(jsonData);

        // find longest array in original container
        JSONArray longest = null;
        for (int i=0; i<array.length(); i++) {
            JSONArray a = array.getJSONArray(i);
            if (longest == null || a.length() > longest.length()) {
                longest = a;
            }
        }

        // build a map with the people in "longest", for quick lookup
        HashMap<String, Integer> bosses = new HashMap<String, Integer>();
        for (int i=0; i<longest.length(); i+=2) {
            bosses.put(longest.getString(i) + "|" + longest.getString(i+1), i);
        }

        // prepare target container       
        ArrayList<JSONArray> container = new ArrayList<JSONArray>();

        // fill in missing values
        for (int i=0; i<array.length(); i++) {
            JSONArray a = array.getJSONArray(i);
            ArrayList<String> refactored = new ArrayList<String>();
            // copy leaf employee
            refactored.add(a.getString(0));
            refactored.add(a.getString(1));
            for (int j=2; j<a.length(); j+=2) {
                // possibly fill in nulls before adding this boss
                String boss = a.getString(j) + "|" + a.getString(j+1);
                if (bosses.containsKey(boss)) {
                    for (int k=j; k<bosses.get(boss); k++) {
                        // pad with nulls until we reach target position
                        refactored.add(null);
                    }
                }
                refactored.add(a.getString(j));
                refactored.add(a.getString(j+1));
            }
            container.add(new JSONArray(refactored));
        }
        return new JSONArray(container).toString();
    }

    public static void main(String args[]) {
        System.out.println(refactor(args[0]));
    }
}