LintCode Heapify

原题链接在这里：http://www.lintcode.com/en/problem/heapify/

题目：

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.

题解：

从array的中段开始向前对每一个点做sift down.

Time Complexity: O(n). Proof link: http://stackoverflow.com/questions/9755721/how-can-building-a-heap-be-on-time-complexity

Space: O(1).

AC Java:

public class Solution {
    /**
     * @param A: Given an integer array
     * @return: void
     */
    public void heapify(int[] A) {
        if(A == null || A.length == 0){
            return;
        }
        for(int i = A.length/2; i>=0; i--){
            siftDown(A, i);
        }
    }
    
    private void siftDown(int [] A, int i){
        int len = A.length;
        while(i < len){
            int smallIndex = i;
            if(2*i+1 < len && A[2*i+1] < A[smallIndex]){
                smallIndex = 2*i+1;
            }
            if(2*i+2 < len && A[2*i+2] < A[smallIndex]){
                smallIndex = 2*i+2;
            }
            if(smallIndex == i){
                break;
            }
            swap(A, i, smallIndex);
            i = smallIndex;
        }
    }
    
    private void swap(int [] nums, int i, int j){
        int temp = nums[i];
        nums[i] = nums[j];
        nums[j] = temp;
    }
}