Data structure - pile - heap heap of maintenance -130.

2020-04-01 08:15:56

Problem Description:

Given an array of integers, the stack operation is to put it into a minimum heap array.

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

Sample

输入 : [3,2,1,4,5]
输出 : [1,2,3,4,5]
解释 : 返回任何一个合法的堆数组

challenge

O (n) time complexity of the complete stack

Problem Solving:

Since the stack is its nature is a complete binary tree, then we can begin filtering downward from the non-leaf nodes, the specific operation from above is set to the minimum of the root.

By estimating the number of nodes per filter, i.e., n / 2 + n / 4 + ... + 1 = n.

Time complexity: O (n)

    public void heapify(int[] A) {
        for (int i = A.length / 2; i >= 0; i--) {
            siftdown(A, i);
        }
    }
    
    private void siftdown(int[] A, int idx) {
        if (idx >= A.length) return;
        int k = idx;
        if (idx * 2 + 1 < A.length && A[idx * 2 + 1] < A[k]) k = idx * 2 + 1;
        if (idx * 2 + 2 < A.length && A[idx * 2 + 2] < A[k]) k = idx * 2 + 2;
        if (k == idx) return;
        int temp = A[idx];
        A[idx] = A[k];
        A[k] = temp;
        siftdown(A, k);
    }