Question

link1, link2, link3.

There is an array of integers which represent heights of persons.

Given another array… Let’s call it count-array. It contain how many persons in front of him are greater than him in height.

求原数组。(原数组中元素是从1到n。)

Example: 

Input(Count array): 0, 0, 2, 0
Output(原数组): 2, 3, 1, 4

求nlogn的算法。

Solution

This is naive solution from floor 29 of this thread:

总结一下，用一个List存放1…n。

从头到尾扫描给定的数组，每得到一个值，从List里删掉。

因为List里数据是有序的，因此remove操作可以使用二分法，复杂度为O(logn).

这样本算法复杂度为O(nlogn).

Example: 

count array 
i C[0,0,2,0]   N[4, 3, 2, 1]
3 C[3] = 0     在N里面删除N[0]=4, N=[3, 2, 1],   Ans=[4]
2 C[2] = 2     在N里面删除N[2]=1, N=[3, 2],   Ans=[1, 4]
1 C[1] = 0     在N里面删除N[0]=3, N=[2],   Ans=[3, 1, 4]
0 C[0] = 0     在N里面删除N[0]=2, N=[], Ans=[2, 3, 1, 4]

But there is a problm here, since removing item from list requires O(n), we will achieve O(n^2) time. How do we optimize this?

The answer is BST with each node keeping track of how many nodes is on the left branch, and how many on right branch. For details of this type of TreeNode, refer to [CC150v5] 11.8 Get Rank in Stream of Integers.

The conclusion:

可以化归为这样一道题：

设计一个有序的数据结构，最初里头有自然数1到n这n个元素，

随后这些元素可以被删除，但剩下元素仍然保持有序。

要求实现方法GetKthElement(int k)和RemoveKthElemenet(int k)，

使得它们在任意时刻都不超过O(lgN), N为当前的元素个数

感觉要结合BST之类

Code

Naive approach, O(n^2): 

public int[] form(int peopleCount, int[] countArray) {
    int len = peopleCount;
    int[] heightQueue = new int[len];
    List<Integer> list = new ArrayList<Integer>();
    for (int i = peopleCount; i > 0; i--) {
        list.add(i);
    }
    for (int i = len - 1; i >= 0; i--) {
        heightQueue[i] = list.remove(countArray[i]);
    }
    return heightQueue;
}