Heapify

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].

 

Example

Given [3,2,1,4,5], return [1,2,3,4,5] or any legal heap array.

Challenge i

O(n) time complexity

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.

 

public class Solution {
    /**
     * @param A: Given an integer array
     * @return: void
     */
    public void heapify(int[] A) {
        // write your code here
        if(A==null || A.length==0) return;
        
        for(int i=A.length/2-1;i>=0;i--)
        {
            helper(A,i);
        }
        
    }
    
    public void helper(int[] A,int i)
    {
        //int left=2*i+1>=A.length?Integer.MAX_VALUE:A[2*i+1];
        //int right=2*i+2>=A.length?Integer.MAX_VALUE:A[2*i+2];
        int left=Integer.MAX_VALUE;
        int right=Integer.MAX_VALUE;
        
        if(2*i+1<A.length)
        {
            left=A[2*i+1];
        }
        if(2*i+2<A.length)
        {
            right=A[2*i+2];
        }
        
        //if left is the minimum one in left,righ and A[i].swap it with A[i];
        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);
        }
    }
}