// Version Linpz
public class Solution {
/**
* @param A: Given an integer array
* @return: void
*/
private void siftdown(int[] A, int k) {
while (k * 2 + 1 < A.length) {
int son = k * 2 + 1;
if (k * 2 + 2 < A.length && A[son] > A[k * 2 + 2])
son = k * 2 + 2;
if (A[son] >= A[k])
break;
int temp = A[son];
A[son] = A[k];
A[k] = temp;
k = son;
}
}
public void heapify(int[] A) {
for (int i = (A.length - 1) / 2; i >= 0; i--) {
siftdown(A, i);
}
}
}
// Version 1: this cost O(n)
public class Solution {
/**
* @param A: Given an integer array
* @return: void
*/
private void siftdown(int[] A, int k) {
while (k < A.length) {
int smallest = k;
if (k * 2 + 1 < A.length && A[k * 2 + 1] < A[smallest]) {
smallest = k * 2 + 1;
}
if (k * 2 + 2 < A.length && A[k * 2 + 2] < A[smallest]) {
smallest = k * 2 + 2;
}
if (smallest == k) {
break;
}
int temp = A[smallest];
A[smallest] = A[k];
A[k] = temp;
k = smallest;
}
}
public void heapify(int[] A) {
for (int i = A.length / 2; i >= 0; i--) {
siftdown(A, i);
} // for
}
}
// Version 2: This cost O(nlogn)
public class Solution {
/**
* @param A: Given an integer array
* @return: void
*/
private void siftup(int[] A, int k) {
while (k != 0) {
int father = (k - 1) / 2;
if (A[k] > A[father]) {
break;
}
int temp = A[k];
A[k] = A[father];
A[father] = temp;
k = father;
}
}
public void heapify(int[] A) {
for (int i = 0; i < A.length; i++) {
siftup(A, i);
}
}
}