算法导论之最大子数组问题 C语言实现

最新推荐文章于 2023-08-08 20:30:59 发布

原创最新推荐文章于 2023-08-08 20:30:59 发布 · 708 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#算法导论 #最大子数组 #C语言实现

算法导论学习专栏收录该内容

1 篇文章

订阅专栏

本文介绍了一种使用C语言解决最大子数组问题的方法，该问题源自《算法导论》一书。文章详细展示了如何克服C语言中单一返回值限制，并通过递归和分治策略找到数组中具有最大和的连续子数组。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

暑假到了，可以开始整天的沉迷算法啦，敲开心的，开啃《算法导论》！

最大子数组问题是《算法导论》分治策略一章中的引例，其实并不是很难理解，但是，作为准大二的小白，用C语言实现最大子数组还是挺难的。书上的伪代码指明有三个返回值，但是C语言函数最多只能有一个返回值，所以为了解决这个棘手的问题，本问题代码花费了大约四个小时的事件，幸好午睡的时候灵感来了。

本代码在visual studio 2015上编译通过，计算机系统为Windows 10，希望大家喜欢

// ConsoleApplication3.cpp : 定义控制台应用程序的入口点。
//

#include "stdafx.h"
#include<stdio.h>

int flag = 0;
int low_save, high_save, sum_save;
int left_low, left_high, right_low, right_high, left_sum, right_sum, cross_low, cross_high, cross_sum;

void max_crossarray(int arr[], int low, int high)
{
	int i, j;
	int cross_left_sum = -1000, cross_right_sum = -1000, right, left;
	int mid = (low + high) / 2;
	int sum = 0;
	for (i = mid; i >= low; i--)
	{
		sum += arr[i];
		if (sum > cross_left_sum)
		{
			cross_left_sum = sum;
			left = i;
		}
	}
	if (cross_left_sum == -1000)
	{
		cross_left_sum = sum;
		left = low;
	}
	sum = 0;
	for (j = mid + 1; j <= high; j++)
	{
		sum += arr[j];
		if (sum > cross_right_sum)
		{
			cross_right_sum = sum;
			right = j;
		}
	}
	if (cross_right_sum == -1000)
	{
		cross_right_sum = sum;
		right = high;
	}
	cross_low = left;
	cross_high = right;
	cross_sum = cross_left_sum + cross_right_sum;
}

void max_subarray(int arr[], int low, int high, int flag)
{
	if (low == high)
	{
		if (flag == 1)
		{
			left_low = low;
			left_high = high;
			left_sum = arr[low];
		}
		else if (flag == 2)
		{
			right_low = low;
			right_high = high;
			right_sum = arr[high];
		}
		return;
	}
	int mid = (low + high) / 2;
	max_subarray(arr, low, mid, 1);
	max_subarray(arr, mid + 1, high, 2);
	max_crossarray(arr, low, high);
	if (left_sum >= right_sum&&left_sum >= cross_sum)
	{
		if (flag == 1&&left_sum>sum_save)
		{
			low_save = left_low;
			high_save = left_high;
			sum_save = left_sum;
		}
		else if (flag == 2)
		{
			left_low = right_low;
			left_high = right_high;
			left_sum = right_sum;
			if (right_sum > sum_save)
			{
				low_save = left_low;
				high_save = left_high;
				sum_save = left_sum;
			}
		}
		return;
	}
	else if (right_sum >= left_sum&&right_sum >= cross_sum)
	{
		if (flag == 1)
		{
			right_low = left_low;
			right_high = left_high;
			right_sum = left_sum;
			if (left_sum > sum_save) 
			{
				low_save = left_low;
				high_save = left_high;
				sum_save = left_sum;
			}
		}
		else if (flag == 2&&right_sum>sum_save)
		{
			low_save = right_low;
			high_save = right_high;
			sum_save = right_sum;
		}
		return;
	}
	else if (cross_sum >= left_sum&&cross_sum >= right_sum)
	{
		if (flag == 1)
		{
			left_low = cross_low;
			left_high = cross_high;
			left_sum = cross_sum;
			if (cross_sum > sum_save)
			{
				low_save = cross_low;
				high_save = cross_high;
				sum_save = cross_sum;
			}
		}
		else if (flag == 2)
		{
			right_low = cross_low;
			right_high = cross_high;
			right_sum = cross_sum;
			if (cross_sum > sum_save)
			{
				low_save = cross_low;
				high_save = cross_high;
				sum_save = cross_sum;
			}
		}
		return;
	}
}


int main()
{
	//define a string
	int arr[12] = { 11,-12,3,23,4,-7,6,-9,23,7,1,-9};
	max_subarray(arr, 0, 11, 0);
	printf("the string's max subarray is %d ,from %d to %d\n", sum_save, low_save, high_save);
	return 0;
}

运行结果：