Gas Station——又是一道经典问题

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.

1 如果总的gas - cost小于零的话，那么没有解返回-1

2 如果前面所有的gas - cost加起来小于零，那么前面所有的点都不能作为出发点。

3 选择最佳出发点后，判断从这点出发能否环行一圈。

class Solution {
public:

    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
            //vector<int> sum;
            int sum=;
            int res=-;
            for(int i=;i<gas.size();i++)
            {
                int temp=gas[i]-cost[i];
                sum+=temp;
                if(res==-&&sum>=)
                        res=i;
                else if(sum<)
                {
                    sum=;
                    res=-;
                }
            }
            if(res!=-)
            {
                for(int i=;i<res;i++)
                {
                    int temp=gas[i]-cost[i];
                    sum+=temp;
                    if(sum<)
                    {
                          res=-;
                          break;
                    }
                }

            }
            return res;
    }
};