最大和子序列问题

这个问题是算法导论的一个示例，为了讲解分治。

//算法导论中的分治策略版本


#include<iostream>
using namespace std;
int maxCrossSum(int a[], int begin, int mid, int end)
{
    int sumLeft = a[mid];
    int sumNow=0;
    for (int i = mid; i >= begin; --i)
    {
        sumNow += a[i];
        if (sumNow > sumLeft)
            sumLeft = sumNow;
    }
    int sumRight = a[mid + 1];
    sumNow = 0;
    for (int i = mid + 1; i <= end; ++i)
    {
        sumNow += a[i];
        if (sumNow > sumRight)
            sumRight = sumNow;
    }
    return sumLeft + sumRight;
}
int maxSubArray(int a[], int begin, int end)
{    
    if (begin == end)
        return a[begin]; 
    else
    {
        int mid = (begin + end) / 2;
        int leftSum = maxSubArray(a, begin, mid);
        int rightSum = maxSubArray(a, mid + 1, end);
        int crossSum = maxCrossSum(a, begin, mid, end);
        int sum;
        if (leftSum > rightSum)
            sum = leftSum;
        else sum = rightSum;
        if (sum < crossSum)
            sum = crossSum;
        return sum;
    }
}
int main()
{
    const int SIZE = 13;
    int a[SIZE] = { -3, -15, 20, -3, -16, -23, 18, 20, -9, 12, -5, -22, 15 };
    int maxsubarray = maxSubArray(a, 0, 12);
    cout << maxsubarray << endl;
    system("pause");

这里提供一个更加简便的方法：

更加高效的版本，无须递归，O（n）的时间复杂度
#include<iostream>
using namespace std;
int maxSubArray(int a[], int begin, int end)
{    
    int sum = 0;
    int i = 0;
    int maxSum = 0;
    for (int i = begin; i <= end; ++i)
    {    
        sum += a[i];
        if (sum > maxSum)
            maxSum = sum;
        //cout << maxSum << endl;
        if (sum < 0)
            sum = 0;
    }
    
    return maxSum;
}
int main()
{
    const int SIZE = 13;
    int a[SIZE] = { -3, -15, 20, -3, -16, -23, 18, 20, -9, 12, -5, -22, 15 };
    int maxsubarray = maxSubArray(a, 0, 12);
    cout << maxsubarray << endl;
    system("pause");
}  

此处不得不说，后面的方法兼具大气魄，有远见等特点，方能如此简单便捷！