为何基于比较的排序算法无法快于O(n log n)？堆排序相关疑问

Great question! This is one of the foundational results in algorithm theory, and it all boils down to understanding the decision tree model that underpins every comparison-based sorting algorithm. Let’s break this down in plain terms:

1. The Decision Tree Analogy #

Think of any comparison-based sort (like the heap sorts you’re studying) as a binary decision tree. Every internal node in this tree represents a single pairwise comparison (e.g., "Is element X greater than element Y?"). Each leaf node represents one of the possible sorted permutations of your input list.

For a list of n distinct elements, there are exactly n! (n factorial) possible sorted outcomes. That means our decision tree must have at least n! leaf nodes to account for every possible input order.

2. Calculating the Minimum Number of Comparisons #

In a binary tree, the number of leaf nodes is bounded by 2^h, where h is the tree’s height (the longest path from root to leaf, which equals the maximum number of comparisons needed to sort a list). Since we need at least n! leaves, we get this key inequality:

2^h ≥ n!

Taking the base-2 logarithm of both sides rearranges to:

h ≥ log₂(n!)

3. Stirling’s Approximation Seals the Bound #

To make sense of log₂(n!), we use Stirling’s approximation, which simplifies factorials for large n:

n! ≈ n^n * e^(-n) * √(2πn)

When we take the base-2 log of this approximation, the dominant term (the one that grows fastest as n increases) is n log₂ n. All other terms are lower-order and become negligible for large input sizes. This means the minimum height of our decision tree is Ω(n log n) — in other words, no comparison-based sorting algorithm can do better than O(n log n) time in the worst case.

4. A Quick Note on Non-Comparison Sorts #

This lower bound only applies to algorithms that rely solely on pairwise comparisons. Algorithms like counting sort, radix sort, or bucket sort skip direct comparisons entirely by leveraging specific properties of the input (like integer ranges or digit positions), letting them run in O(n) time for certain cases. But these are not general-purpose — they only work for specific data types, whereas comparison-based sorts (like the heap sorts you’re researching) work for any data that can be compared.

内容的提问来源于stack exchange，提问作者Somit