在O(n)的排序双向链表中寻找给定平均值的最长子表,首先需要明确问题的具体要求和解决方案。根据问题描述,可以将问题分解为以下几个步骤来求解。
下面给出一个基本实现的示例代码(使用Python语言):
class Node:
def __init__(self, val):
self.val = val
self.prev = None
self.next = None
def find_longest_sublist(head, target_avg):
# Step 1: 遍历链表,获取节点值
node_vals = []
current = head
while current:
node_vals.append(current.val)
current = current.next
# Step 2: 计算累积和和位置信息
cumulative_sum = [0] * (len(node_vals) + 1)
pos_dict = {0: 0}
for i in range(len(node_vals)):
cumulative_sum[i+1] = cumulative_sum[i] + node_vals[i]
if cumulative_sum[i+1] not in pos_dict:
pos_dict[cumulative_sum[i+1]] = i+1
# Step 3: 查找目标平均值
target_sum = target_avg * len(node_vals)
# Step 4: 查找最长子表
start, end = 0, 0
min_diff = float('inf')
for i in range(1, len(cumulative_sum)):
complement = cumulative_sum[i] - target_sum
if complement in pos_dict:
diff = i - pos_dict[complement]
if diff < min_diff:
min_diff = diff
start = pos_dict[complement]
end = i
# Step 5: 输出结果
if start == end:
return None
else:
return node_vals[start:end]
# 示例用法
# 创建一个双向链表并设置节点值
head = Node(1)
n2 = Node(2)
n3 = Node(3)
n4 = Node(4)
n5 = Node(5)
head.next = n2
n2.prev = head
n2.next = n3
n3.prev = n2
n3.next = n4
n4.prev = n3
n4.next = n5
n5.prev = n4
# 指定目标平均值
target_avg = 3.5
# 调用函数并输出结果
result = find_longest_sublist(head, target_avg)
if result is None:
print("找不到满足条件的子表")
else:
print("给定平均值的最长子表为:", result)
以上示例代码是一个简单的实现,能够在O(n)的时间复杂度内解决问题。但需要注意,实际应用中可能需要考虑更多的边界条件和异常处理,以及进一步的性能优化。这里只提供了一个基本的思路和实现方式,具体的应用场景和需求可以根据实际情况进行调整。
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