Gas station [Leetcode] two solutions

Because the total gas is greater than the total cost of time. You will be able to circle the whole city.

First Solution :

If there is enough oil at the beginning. Departure from position I. The remaining amount of oil at position K is L (i,k).

to the random K. L (i,k) According to the difference of I, only the difference constant.

We just need to find the smallest L (0, K) corresponding to the k,k+1 to be asked.

The code is as follows:

    int Cancompletecircuit (vector<int> &gas, vector<int> &cost) {        int start = 0;        int Curgas = 0, Mingas = 0, Totalgas = 0;        for (int i = 0; i < gas.size (); i++)        {            int temp = gas[i]-cost[i];            Curgas + = temp;            Totalgas + = temp;            if (Mingas > Curgas)            {                start = i + 1;                Mingas = Curgas;            }        }        if (totalgas >= 0) return start% gas.size ();        else return-1;    }

Another solution：

Assuming L (I,k) < 0, the entire position between I and K is not to K

So from the location of the k+1 from 0 to start looking for

    int Cancompletecircuit (vector<int> &gas, vector<int> &cost) {        int start = 0;        int Curgas = 0, Totalgas = 0;        for (int i = 0; i < gas.size (); i++)        {            int temp = gas[i]-cost[i];            Curgas + = temp;            Totalgas + = temp;            if (Curgas < 0)            {                start = i + 1;                Curgas = 0;            }        }        if (totalgas >= 0) return start% gas.size ();        else return-1;    }

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Gas station [Leetcode] two solutions