Gas Station

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].
You have a car with an unlimited gas tank and it costs cost[i] of
gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.
Return the starting gas station's index if you can travel around the circuit once, otherwise return -1.
Note:

The solution is guaranteed to be unique.
遍历向量，同时计算总的gas-cost，如果小于0，说明不可能存在；
如果大于等于0，需要计算当前为止的gas-cost的和，如果小于0，说明可能的点发生在接下来的station。

int canCompleteCircuit(vector<int> &gas, vector<int> &cost)
{
int temp=0,sum=0,start=0;
int station_num = gas.size();
for(int i=0; i<station_num; i++)
{
temp+=gas[i]-cost[i];
sum+=gas[i]-cost[i];
if(temp < 0)
{
start=(i+1)%station_num;
temp=0;
}
}

if(sum <0)
return -1;
else
return start;
}