我仍在努力理解BigO表示法和时间复杂性，但我真的不确定这个算法(我的代码)的时间复杂性是多少。
// 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("%0.1fn", 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;
}
我不确定它是否是O(N)，但我不这么认为，因为第二个for循环中有一个嵌套循环。然而，它可能不是O(N^2)也。。。更像O((N^2)/2)。
有人能给我建议吗？
算法的复杂度为O(n + m)，其中n是输入的大小，m是"峰值"的数量，无论这些是什么。
原因是核心算法由两个循环组成，大约运行n次。其中一个循环包含一个内部循环。我们需要计算内部循环的主体执行了多少次。
内部循环是在遇到峰值时运行的，看起来循环的主体执行的总次数大致等于峰值的数量。循环嵌套并不重要：对于复杂性计算，主体的总迭代次数才是最重要的。
(通常嵌套循环的迭代次数是相乘而不是相加的，因为它在外循环的每次迭代中都是完全执行的，但这里的情况并非如此。(在内循环中)从第一个峰值到最后一个峰值进行逻辑迭代；请注意，在外部循环的迭代之间，要跟踪(使用p)break在内部循环之外的位置，并在返回内部循环时从p开始。)
我遗漏了什么吗？我宁愿您只提供了伪代码/或必要的代码行，但看起来一行只有两个for循环，第二个for循环有一个嵌套循环。那将是O(N^2)，不是吗？随着数据集的增加，时间复杂性应该与数据集的平方成比例。。。
不过，如果有人想纠正我的话，我从来没有掌握过这个科目
编辑--请参阅注释以了解错误的原因。。！