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Heapify.py
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84 lines (66 loc) · 2.88 KB
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"""
题目描述：给出一个整数数组，堆化操作就是把它变成一个最小堆数组。对于堆数组A，A[0]是堆的根，
并对于每个A[i]，A [i * 2 + 1]是A[i]的左儿子并且A[i * 2 + 2]是A[i]的右儿子。
样例：给出 [3,2,1,4,5]，返回[1,2,3,4,5] 或者任何一个合法的堆数组
"""
class Solution:
# @param A: Given an integer array
# @return: void
def heapify(self, A):
for i in A:
cur = i
par = (i - 1) // 2
while par >= 0 and A[par] > A[cur]:
A[par], A[cur] = A[cur], A[par]
cur = par
par = (cur-1)//2
"""
heap:
class Heap(object):
def __init__(self, elements):
self.vector = []
for element in elements:
self.add(element)
def add(self, element):
# 添加新元素到列表末尾
self.vector.append(element)
# 找到新添加节点的父亲
cur = len(self.vector) - 1
par = (cur - 1) // 2
# 逐层交换
while cur != 0 and self.vector[cur] > self.vector[par]:
self.vector[par], self.vector[cur] = self.vector[cur], self.vector[par]
cur = par
par = (cur - 1) // 2
def pop(self):
# Next决定是否还需要进行下一步交换
Next = True
# 首尾交换
self.vector[0], self.vector[-1] = self.vector[-1], self.vector[0]
# 将尾元素（其实是根节点）删除出来
result = self.vector.pop()
cur = 0
# 我们认为只要这一步进行交换了，且还能交换（当前节点还有孩子），就继续交换
while cur < len(self.vector) and Next:
Next = False
# 找左右孩子的索引
left_child, right_child = (2 * cur) + 1, (2 * cur) + 2
# 左孩子索引越界，交换终止
if left_child >= len(self.vector):
break
# 右孩子存在
if right_child < len(self.vector):
# max_index：较大孩子的索引
max_index = right_child if self.vector[left_child] < self.vector[right_child] else left_child
if self.vector[cur] < self.vector[max_index]:
self.vector[cur], self.vector[max_index] = self.vector[max_index], self.vector[cur]
cur = max_index
Next = True
# 右孩子不存在
elif self.vector[cur] < self.vector[left_child]:
self.vector[cur], self.vector[left_child] = self.vector[left_child], self.vector[cur]
cur = left_child
Next = True
# 返回被删除的值
return result
"""