LintCode "Continuous Subarray Sum II"

Flip over your mind: in rotated subarray case, we can simply cut the continuous smallest subarray.

class Solution {
public:
    /**
     * @param A an integer array
     * @return  A list of integers includes the index of
     *          the first number and the index of the last number
     */

    vector<int> continuousSubarraySumII(vector<int>& A) {
        vector<int> ret;

        size_t len = A.size();

        long long sum = A[0], csum = A[0];
        int s = 0, e = 0, ss = 0, ee = 0; // for max continuous

        long long sum0= A[0], csum0= A[0];
        int s0= 0, e0= 0, ss0= 0, ee0= 0; // for min continuous

        bool bAllNeg = true;
        for (int i = 1; i < len; i++)
        {
            if(A[i] >= 0) bAllNeg = false;
            //  max
            long long nsum = csum + A[i];
            if (A[i] > nsum)
            {
                ss = ee = i;
                csum = A[i];
            }
            else
            {
                ee = i;
                csum = nsum;
            }

             if(csum > sum)
            {
                sum = csum;
                s = ss;
                e = ee;
            }

            //  min
            long long nsum0 = csum0 + A[i];
            if (A[i] < nsum0)
            {
                ss0 = ee0 = i;
                csum0= A[i];
            }
            else
            {
                ee0 = i;
                csum0= nsum0;
            }
            if(csum0 < sum0)
            {
                sum0 = csum0;
                s0 = ss0;
                e0 = ee0;
            }
        }

        long long asum = accumulate(A.begin(), A.end(), 0);
        long long osum = asum - sum0;
        if (bAllNeg)
        {
            int inx = max_element(A.begin(), A.end()) - A.begin();
            ret.push_back(inx);
            ret.push_back(inx);
        }
        else if (sum >= osum)
        {
            ret.push_back(s);
            ret.push_back(e);
        }
        else
        {
            ret.push_back(e0 + 1);
            ret.push_back(s0 - 1);
        }
        return ret;
    }
};