最大子序列和问题

问题描述：给定整数A1,A2,...An（可能有负数），求连续项和的最大值。例如对于输入-2,11，-4,13，-5，-2，答案为20（A2到A4）。

O(N³)

int maxSubSum1(const vector<int>& a)
{
    int maxSum = 0;
    for (int i = 0; i < a.size(); i++)
    {
        for (int j = i; j < a.size(); j++)
        {
            int thisSum = 0;
            for (int k = i; k <= j; k++)
            {
                thisSum += a[k];
            }
            if (thisSum > maxSum)
                maxSum = thisSum;
        }
    }
    return maxSum;
}

O(N²)

int maxSubSum2(const vector<int>& a)
{
    int maxSum = 0;
    for (int i = 0; i < a.size(); i++)
    {
        int thisSum = 0;
        for (int j = i; j < a.size(); j++)
        {
            thisSum += a[j];
            if (thisSum > maxSum)
                maxSum = thisSum;
        }
    }
    return maxSum;
}

O(NlgN) 采用分治策略

int max3(int a, int b, int c)
{
    int temp = a > b ? a : b;
    return temp > c ? temp : c;
}
int maxSumRec(const vector<int>& a, int left, int right)
{
    if (left == right)
    {
        if (a[left] > 0)
            return a[left];
        else
            return 0;
    }
    int center = (left + right) / 2;
    int maxLeftSum = maxSumRec(a, left, center);
    int maxRightSum = maxSumRec(a, center + 1, right);
    int maxLeftBorderSum = 0,leftBorderSum=0;
    for (int i = center; i >= left; i--)
    {
        leftBorderSum += a[i];
        if (leftBorderSum > maxLeftBorderSum)
            maxLeftBorderSum = leftBorderSum;
    }
    int maxRightBorderSum = 0, rightBorderSum = 0;
    for (int j=center+1;j<=right;j++)
    {
        rightBorderSum += a[j];
        if (rightBorderSum > maxRightBorderSum)
            maxRightBorderSum = rightBorderSum;
    }
    return max3(maxLeftSum, maxRightSum, maxLeftBorderSum + maxRightBorderSum);
}
int maxSubSum3(const vector<int>& a)
{
    return maxSumRec(a, 0, a.size() - 1);
}

 O(N)

int maxSubSum4(const vector<int>& a)
{
    int maxSum = 0, thisSum = 0;
    for (int j=0;j<a.size();j++)
    {
        thisSum += a[j];
        if (thisSum > maxSum)
            maxSum = thisSum;
        else if (thisSum < 0)
            thisSum = 0;
    }
    return maxSum;
}