Lintcode: 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].
基础题，一定要熟练掌握

public class Solution {
    public void heapify(int[] A) {
        for (int i = A.length / 2; i >= 0; i--) {
        	heapify(A, i);
        }
    }

    private void heapify(int[] A, int i) {
    	if (i > A.length) {
    		return;
    	}
    	int left = i * 2 + 1 < A.length ? A[i*2+1] : Integer.MAX_VALUE;
    	int right = i * 2 + 2 < A.length ? A[i*2+2] : Integer.MAX_VALUE;
    	if (left < right && left < A[i]) {
    		A[i*2+1] = A[i];
    		A[i] = left;
    		heapify(A, i*2+1);
    	} else if (right <= left && right < A[i]) {
    		A[i*2+2] = A[i];
    		A[i] = right;
    		heapify(A, i*2+2);
    	}
    }
}