[LintCode] Heapify

Problem

Given an integer array, heapify it into a min-heap array.

For a heap array A, A[0] is the root of heap, and for each A[i], A[i * 2 + 1] is the left child of A[i] and A[i * 2 + 2] is the right child of A[i].

Clarification

What is heap?
Heap is a data structure, which usually have three methods: push, pop and top. where "push" add a new element the heap, "pop" delete the minimum/maximum element in the heap, "top" return the minimum/maximum element.

What is heapify?
Convert an unordered integer array into a heap array. If it is min-heap, for each element A[i], we will get A[i * 2 + 1] >= A[i] and A[i * 2 + 2] >= A[i].

What if there is a lot of solutions?
Return any of them.
Example
Given [3,2,1,4,5], return [1,2,3,4,5] or any legal heap array.

Challenge

O(n) time complexity

Note

首先，先搞懂几个概念：heap，min-heap，complete tree。这里只要知道heap是一种complete tree的树结构，结点i的左右子结点的index为2*i+1和2*i+2，min-heap是子结点大于根节点的树，就大概明白题目要怎么heapify了。
思路是从底部开始构建数组，将较小的结点移向上层结点。先取数组末尾的两个元素为left和right，若2*i+1和2*i+2越界，则赋Integer.MAX_VALUE，这样可以在后面的分支语句避免对越界情况不必要的讨论。然后，取left和right的较小者和上层A[i]结点交换，实现min-heap结构。交换了A[i]和A[2*i+1]/A[2*i+2]之后，还要重新heapify被交换到末位的原先的A[i]，即交换之后的A[2*i+1]/A[2*i+2]。然后i不断减小，直到整个数组完成heapify。
其实，这就相当于对数组A进行堆排序。

Arrays.sort(A);

Solution

public class Solution {
    public void heapify(int[] A) {
        for (int i = A.length/2; i >= 0; i--) {
            helper(A, i);
        }
    }
    public void helper(int[] A, int i) {
        if (i > A.length) return;
        int left = 2*i+1 < A.length ? A[2*i+1] : Integer.MAX_VALUE;
        int right = 2*i+2 < A.length ? A[2*i+2] : Integer.MAX_VALUE;
        if (left < right && left < A[i]) {
            A[2*i+1] = A[i];
            A[i] = left;
            helper(A, 2*i+1);
        }
        else if (right < A[i]) {
            A[2*i+2] = A[i];
            A[i] = right;
            helper(A, 2*i+2);
        }
    }
}