該算法（Big-O）的時間復雜度是多少？

Question

我仍在嘗試了解BigO表示法和時間復雜度，但是，我真的不確定此算法（我的代碼）的時間復雜度是多少。

// 03_BeaverConstructions.c
// Created for FIKS on 28/12/2013 by Dominik Hadl
//
// Time complexity: O(N+M)
// Space complexity: O(N)
// ------------------------------------------
// LICENSE (MIT)
// Copyright (c) 2013 Dominik Hadl
//
// Permission is hereby granted, free of charge, to any person obtaining a copy
// of this software and associated documentation files (the "Software"), to deal
// in the Software without restriction, including without limitation the rights
// to use, copy, modify, merge, publish, distribute, sublicense, and/or sell
// copies of the Software, and to permit persons to whom the Software is
// furnished to do so, subject to the following conditions:
//
// The above copyright notice and this permission notice shall be included in
// all copies or substantial portions of the Software.
//
// THE SOFTWARE IS PROVIDED "AS IS", WITHOUT WARRANTY OF ANY KIND, EXPRESS OR
// IMPLIED, INCLUDING BUT NOT LIMITED TO THE WARRANTIES OF MERCHANTABILITY,
// FITNESS FOR A PARTICULAR PURPOSE AND NONINFRINGEMENT. IN NO EVENT SHALL THE
// AUTHORS OR COPYRIGHT HOLDERS BE LIABLE FOR ANY CLAIM, DAMAGES OR OTHER
// LIABILITY, WHETHER IN AN ACTION OF CONTRACT, TORT OR OTHERWISE, ARISING FROM,
// OUT OF OR IN CONNECTION WITH THE SOFTWARE OR THE USE OR OTHER DEALINGS IN
// THE SOFTWARE.
// ------------------------------------------

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

// ------------------------------------------
// Setup
// ------------------------------------------

#define MAX_PROFILE_LENGTH 10000
#define NO_VALUE -99999

#define BLOCK_VOLUME 1
#define SLOPE_VOLUME 0.5

const char kSlopeRise = '/';
const char kSlopeLower = '\\';
const char kSlopeStay = '_';

// ------------------------------------------
// Structs
// ------------------------------------------

typedef struct
{
    int start_elevation;
    float current_volume;
} Lake;

typedef struct
{
    int location;
    int elevation;
} Peak;

// ------------------------------------------
// Declarations
// ------------------------------------------

int main(int argc, char const *argv[]);
float get_water_volume_of_profile(char const *hill_profile);

// ------------------------------------------
// Main
// ------------------------------------------

int main(int argc, char const *argv[])
{
    // Get the profile
    char hill_profile[MAX_PROFILE_LENGTH + 1];
    fgets(hill_profile, MAX_PROFILE_LENGTH + 1, stdin);

    // Calculate the volume
    float volume = get_water_volume_of_profile(hill_profile);

    // Print it!
    printf("%0.1f\n", volume);

    return 0;
}

// ------------------------------------------
// Calculation
// ------------------------------------------

float get_water_volume_of_profile(char const *hill_profile)
{
    float total_volume = 0; 
    int current_elevation = 0, number_of_peaks = 0, last_peak_index = 0;

    // Get the actual length of the hill profile    
    int profile_length = strlen(hill_profile);

    // Prepare the peaks and lakes in the hill profile
    Peak peaks[profile_length / 2];
    Lake lake = {NO_VALUE, 0};

    // First, get all the peaks
    for (int i = 0; i < profile_length; i++)
    {
        char current_char = hill_profile[i];
        char next_char = hill_profile[i + 1];

        switch (current_char)
        {
            case kSlopeRise:
                current_elevation += 1;
                break;
            case kSlopeLower:
                current_elevation -= 1;
                break;
            case kSlopeStay:
                break;
        }

        if (next_char == '\n')
        {
            peaks[number_of_peaks].location = i + 1;
            peaks[number_of_peaks].elevation = current_elevation;
            number_of_peaks++;
            break;
        }

        if (current_char == kSlopeRise &&
            (next_char == kSlopeLower || next_char == kSlopeStay))
        {
            peaks[number_of_peaks].location = i + 1;
            peaks[number_of_peaks].elevation = current_elevation;
            number_of_peaks++;
        }
    }

    // Now, go through the profile and get the water volume
    current_elevation = 0;
    for (int i = 0; i < profile_length; i++)
    {
        // Get current char and decide what to do
        char current_char = hill_profile[i];
        switch (current_char)
        {
            case kSlopeRise:
            {
                if (lake.start_elevation != NO_VALUE &&
                    lake.start_elevation > current_elevation)
                {
                    lake.current_volume += SLOPE_VOLUME;
                }

                // Increase the elevation
                current_elevation++;

                if (lake.start_elevation == current_elevation)
                {
                    total_volume += lake.current_volume;

                    lake.start_elevation = NO_VALUE;
                    lake.current_volume = 0;
                    break;
                }

                if (lake.start_elevation != NO_VALUE)
                {
                    int elevation_diff = abs(lake.start_elevation - current_elevation);
                    if (elevation_diff > 1)
                    {
                        lake.current_volume += (elevation_diff - 1) * BLOCK_VOLUME;
                    }
                }

                break;
            }
            case kSlopeLower:
            {
                current_elevation--; // Lower the elevation

                // Set elevation where water starts if not already set
                if (lake.start_elevation == NO_VALUE)
                {
                    for (int p = last_peak_index; p < number_of_peaks; p++)
                    {
                        if (peaks[p].elevation >= current_elevation + 1 &&
                            peaks[p].location > i)
                        {
                            lake.start_elevation = current_elevation + 1;
                            last_peak_index = p;
                            break;
                        }
                    }
                    if (lake.start_elevation == NO_VALUE) 
                    {
                        break;
                    }
                }

                lake.current_volume += SLOPE_VOLUME;

                int elevation_diff = abs(lake.start_elevation - current_elevation);
                if (elevation_diff > 1)
                {
                    lake.current_volume += elevation_diff * BLOCK_VOLUME;
                }

                break;
            }
            case kSlopeStay:
            {
                if (lake.start_elevation != NO_VALUE)
                {
                    int elevation_diff = abs(lake.start_elevation - current_elevation);
                    lake.current_volume += elevation_diff * BLOCK_VOLUME;
                }   
                break;
            }
        }
    }

    // Return the total water volume
    return total_volume;
}

我不確定是否為O（N），但我不這樣認為，因為第二個for循環中有一個嵌套循環。 但是，它也可能不是O（N ^ 2）……更像是O（（N ^ 2）/ 2）。

有人可以給我建議嗎？

Answer 1

算法的復雜度為O(n + m) ，其中n是輸入的大小， m是“峰值”的數目，無論這些數目是多少。

原因是核心算法由大約運行n次的兩個循環組成。 循環之一包含一個內部循環。 我們需要計算內部循環的主體執行了多少次。

當您遇到一個峰時，將運行內部循環，看起來循環主體執行的總次數大約是您擁有的峰數。 循環嵌套並不重要：對於復雜度計算，主體的迭代總數才是最重要的。

（通常，嵌套循環的迭代計數是相乘而不是相加，因為它是在外循環的每次迭代中完全執行，但是這里不是這種情況。您在邏輯上是從第一個峰到最后一個（在內部循環中）進行迭代；請注意，您要跟蹤（使用p ）在外部循環的兩次迭代之間break內部循環的位置，並在返回內部循環時從p開始。）

Answer 2

我想念什么嗎？ 我希望您只提供了偽代碼/或必要的代碼行，但看起來它只是一行中的兩個for循環，而第二個for循環具有嵌套循環。 那將是O（N ^ 2），不是嗎？ 隨着數據集的增加，時間復雜度應與數據集的平方成正比。

但是，如果有人要糾正我，我永遠不會精通此主題。

編輯-請參閱注釋以了解為什么這是錯誤的..！