Maximum sub-segment and maximum stock purchase and sale plan

[-2,2,3,-1]的最大字段和是[2,3]/*动态规划算法:**b[j]=max{a[i]+ +a[j]},1<=i<=j,且1<=j<=n,则所求的最大子段和为max b[j]，1<=j<=n。**由b[j]的定义可易知，当b[j-1]>0时b[j]=b[j-1]+a[j]，否则b[j]=a[j]。故b[j]的动态规划递归式为:**b[j]=max(b[j-1]+a[j],a[j])，1<=j<=n。**T(n)=O(n)*/int MaxSum_DYN(int *v,int n){    int sum=0,b=0;    int i;    for (i=1;i<=n;i++)    {        if(b>0)            b+=v[i];        else            b=v[i];        if(b>sum)            sum=b;    }    return sum;}

 

 

There is a stock problem:

Given an int array with a length of N, the array stores the price of a stock for N days, seeking for the most profitable Purchase and SaleOut of the solution, such as [, 6], then when you buy 7 at 2, you can make the maximum profit.

for(i = 0; i < n; ++i){    minPrice = min(minPrice, a[i]);    best  = max(a[i] - minPrice, best);}

Or the price change array is [-2, 3, 2,-1], followed by the largest sub-segment and question.

 

Maximum sub-segment and maximum stock purchase and sale plan