【算法导论笔记】最大连续子序列和

分别采用两种方法实现：

1. 分而治之递归解法
2. 线性数学解法

1. 分而治之递归解法

分为三种情况，

1. 最大子序列和完全在左子数组；
2. 最大子序列和完全在右子数组；
3. 跨越了中间；

关键代码

/* 通过初始化值，完美规避，必须包含 mid 的子序列 */
int findMaxCrossingSubArray(int * A, int low, int mid, int high)
{
	int leftIndex = mid;
	int rightIndex = mid;
	int leftSum = 0;
	int rightSum = 0;
	int leftFinalSum = A[mid];
	int rightFinalSum = 0;


	for (int i = mid; i >= 0; i--)
	{
		leftSum += A[i];
		if (leftSum > leftFinalSum)
		{
			leftIndex = i;
			leftFinalSum = leftSum;
		}
	}

	for (int i = mid + 1; i < high; i++)
	{
		rightSum += A[i];
		if (rightSum > rightFinalSum)
		{
			rightIndex = i;
			rightFinalSum = rightSum;
		}
	}

	return leftFinalSum + rightFinalSum;
}

int findMaximumSubarray(int *A, int low, int high)
{
	if (low == high)
	{
		return A[low];
	}
	// 注意 mid 的计算方法
	int mid = (high + low) / 2;
	int left = findMaximumSubarray(A, low, mid);
	int right = findMaximumSubarray(A, mid + 1, high);
	int cross = findMaxCrossingSubArray(A, low, mid, high);
	
	return max(max(left, right), cross);
}

2. 数学思路

如果前面加和为负数，我就选择不加，如果是正数我就加入，最终每次加入都需要和之前的结果比较取一个大值返回！参考 int findMaximumSubarray2(int *A, int size);

关键代码

int findMaximumSubarray2(int *A, int size)
{
	int sum = 0;
	int result = A[0];
	for (int i = 0; i < size; i++)
	{
		if (sum < 0)
		{
			sum = 0;
		}

		sum += A[i];
		result = result > sum ? result : sum;
	}
	return result;
}

3. 总结

方法 2 的时间复杂度较低，方法 2 更多的是人为的做了分析取舍，精进了算法，导致问题变得简单。
学而不思则罔，学会了计算机的技能之后，再加上数学的思维，往往会事半功倍！

完整的测试验证代码：

#include "pch.h"
#include <iostream>

#define max(a, b) a > b ? a : b

int findMaxCrossingSubArray(int * A, int low, int mid, int high);
int findMaximumSubarray(int *A, int low, int high);
int findMaximumSubarray2(int *A, int size);

int main()
{
	int a[] = { 13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7 };
	std::cout << findMaximumSubarray(a, 0, 15) << std::endl;
	std::cout << findMaximumSubarray2(a, sizeof(a) / sizeof(a[0])) << std::endl;
	return 0;
}

/* 通过初始化值，完美规避，必须包含 mid 的子序列 */
int findMaxCrossingSubArray(int * A, int low, int mid, int high)
{
	int leftIndex = mid;
	int rightIndex = mid;
	int leftSum = 0;
	int rightSum = 0;
	int leftFinalSum = A[mid];
	int rightFinalSum = 0;


	for (int i = mid; i >= 0; i--)
	{
		leftSum += A[i];
		if (leftSum > leftFinalSum)
		{
			leftIndex = i;
			leftFinalSum = leftSum;
		}
	}

	for (int i = mid + 1; i < high; i++)
	{
		rightSum += A[i];
		if (rightSum > rightFinalSum)
		{
			rightIndex = i;
			rightFinalSum = rightSum;
		}
	}

	return leftFinalSum + rightFinalSum;
}

int findMaximumSubarray(int *A, int low, int high)
{
	if (low == high)
	{
		return A[low];
	}
	// 注意 mid 的计算方法
	int mid = (high + low) / 2;
	int left = findMaximumSubarray(A, low, mid);
	int right = findMaximumSubarray(A, mid + 1, high);
	int cross = findMaxCrossingSubArray(A, low, mid, high);
	
	return max(max(left, right), cross);
}

int findMaximumSubarray2(int *A, int size)
{
	int sum = 0;
	int result = A[0];
	for (int i = 0; i < size; i++)
	{
		if (sum < 0)
		{
			sum = 0;
		}

		sum += A[i];
		result = result > sum ? result : sum;
	}
	return result;
}