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].

Example

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

Challenge

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].

思路：从下往上heapify，如果两个子节点都比母节点小，那么就swap最小的那个。swap之后，我们要重新heapify 被换的节点（曾经的母节点，现在的子节点）

public void heapify(int[] A) {
for (int i = A.length / 2-1; i >= 0; i--) {
helper(A, i);
}
}

void helper(int[] A, int i) {
int left = i * 2 + 1 >= A.length ? Integer.MAX_VALUE : A[2*i+1];
int right = i * 2 + 2 >= A.length ? Integer.MAX_VALUE : A[2*i+2];

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);
}
}