题目：

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.

思路：

思路1：通过两层循环，依次从某个点出发并测试是否能够运行一圈。时间复杂度为O（n2），不满足要求。 思路2：首先确认gas总和大于cost，因此判断能够绕圈。接下来寻找起始位置，我们可以借鉴归并排序的思路，如果某一段路gas>cost，则这段路剩余的油量可以支撑其他路段。因此问题变化为找到某个节点，在它之前的路段剩余油量为负，而从它开始到整个队列结束剩余油量为正（正油量可以不足前面路段的不足油量）。时间可以在O（n）完成。

代码：

思路1：

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        for(int j=0;j<gas.size();j++)
        {
            int tank = 0;
            bool work = true;
            int i=j;
            do
            {
                tank+=gas[i];
                tank-=cost[i];
                if(tank < 0)
                {
                    work = false;
                    break;
                }
                i=(i+1)%gas.size();
            }while(i!=j);
            if(work)
            {
                return j;
            }
        }
        return -1;
    }
};

思路2：

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        vector<int> remainder;
        int sum =0;
        for(int i = 0; i < gas.size(); i++)
        {
            remainder.push_back(gas[i]-cost[i]);
            sum += gas[i]-cost[i];
        }
        if(sum < 0)
        {
            return -1;
        }
        else
        {
            int start;
            int cur = 0;
            do
            {
                start = cur;
                int tmp = remainder[cur++];
                while(tmp >= 0 && cur<gas.size())
                {
                    tmp += remainder[cur++];
                    if(tmp < 0)
                    {
                        break;
                    }
                }
                if(tmp >= 0 && cur == gas.size())
                {
                    return start;
                }
            }while(cur<gas.size());
            return -1;
        }
    }
};