由如何计算一组简单的低买高卖数据获得最大收益引出的最大子数组计算算法。

1. 概述

按照惯例，注释掉的代码可以显示每步的运算结果。
有个地方我是不甚了解，就是下面在代码后面做了注释的部分。
这两个部分本来可以直接 int data = {...};
然后 return data;
但是这样会抛出警告

**warning: ** address of stack memory associated with local variable 'data' returned [-Wreturn-stack-address]

2. 代码示例

return data;
#include <stdio.h>
// int c=0;
void printArray(int* array, int length) {
    printf("\n");
    for(int i=0; i<length;i++) {
        printf("%d ", array[i]);
    }
    printf("\n");

}

int* findMaxCrossingSubarray(int* a, int low, int mid, int high) {

    int leftSum=0x80000000;
    int sum=0;
    int maxLeft = mid;
    for (int i=mid; i>=low; i--) {
        sum+=a[i];
        if (sum>leftSum) {
            leftSum = sum;
            maxLeft = i;
        }
    }
    int rightSum=0x80000000;
    sum=0;
    int maxRight = mid;
    for (int i=mid+1; i<=high; i++) {
        sum+=a[i];
        if (sum>rightSum) {
            rightSum = sum;
            maxRight = i;
        }
    }
    int *data; // ---
    int dataTemp[3] = {maxLeft, maxRight, leftSum+rightSum}; // ---
    data = dataTemp; // ---
    return data;
}
int* findMaxSubarray(int* a, int low, int high) {

    if(high==low) {
        int *data; // ---
        int dataTemp[3] = {low,high,a[low]}; // ---
        data = dataTemp; // ---
        return data;
    } else {
  
        int mid = (low+high)/2;
        int *dataLeft = findMaxSubarray(a, low, mid);
        // printf("这是第%d次递归", c++);
        // printArray(dataLeft, 3);

        int *dataRight = findMaxSubarray(a, mid+1, high);
        int *dataCross = findMaxCrossingSubarray(a, low, mid, high);
        if (dataLeft[2]>dataRight[2]&&dataLeft[2]>dataCross[2]) {
            return dataLeft;
        } else if (dataRight[2]>dataLeft[2]&&dataRight[2]>dataCross[2]) {
            return dataRight;
        } else {
            return dataCross;
        }
    }

}

int main()
{
    int a[] = {1,-2,3,4,5,9};
    int* data = findMaxSubarray(a, 0, sizeof(a)/sizeof(int)-1);
    printArray(data, 3);
    return 0;

}