排列生成算法分析：C++程序时间复杂度解析问询

Great question! Let’s break down the time complexity of your permutation generator program step by step to clarify where those O(n*n!) and O(n) terms come from.

1. Initial Sorting Step #

First, you call std::sort(myints, myints+3) to start with the lexicographically smallest permutation. For an array of size n, std::sort runs in O(n log n) time. That said, once n is even moderately large, n! grows exponentially faster than n log n, so this step becomes negligible compared to the rest of the program’s runtime.

2. The Do-While Loop: Generating and Printing Permutations #

This loop runs exactly n! times—once for every unique permutation of your n elements. Let’s split this into two key parts: printing the permutation and executing std::next_permutation.

a. Printing Each Permutation #

Each iteration prints n elements (like myints[0], myints[1], myints[2] for n=3). Printing a single element is O(1), so printing all n elements takes O(n) time per loop iteration. Multiply that by n! iterations, and you get a total of O(n * n!) time spent on printing.

b. std::next_permutation Calls #

Let’s unpack how std::next_permutation works and its cost:

• Core Logic: The function finds the next lexicographical permutation by:

  1. Scanning from the end of the array to find the first element that can be swapped to create a larger permutation (worst-case O(n) for this scan).
  2. Finding the smallest element larger than that element to swap with (another worst-case O(n) scan).
  3. Swapping those two elements (O(1)).
  4. Reversing the suffix after the swapped position to get the smallest possible permutation starting with the new prefix (worst-case O(n)).

• Total Cost Across All Calls: While a single next_permutation call can take O(n) time (e.g., when reversing almost the entire array), the total time for all n! calls is O(n * n!). This is because each element is involved in a limited number of reversals and swaps across all permutations, and the cumulative operations scale linearly with the total number of elements processed across all permutations (which is n * n!).

3. Overall Time Complexity #

Putting it all together:

• Sorting: O(n log n)
• Printing: O(n * n!)
• next_permutation total: O(n * n!)

Since n * n! completely dominates n log n (even for n=3, n! is 6, and it grows exponentially), the overall time complexity of the program is O(n * n!).

The O(n) you referenced likely refers to the per-iteration cost of printing (or the worst-case per-call cost of next_permutation), but the total runtime is driven by repeating these O(n) operations n! times, leading to the O(n*n!) total.

内容的提问来源于stack exchange，提问作者RyanP