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.

通过gas - cost 构造一个数组left，left[i]表征 i  站到达下一站之后的剩余gas量；

注意点：1.记得把pos取模，从超出位置拉回来就好； 2. 根据聚合分析，时间复杂度是O(N)，因为在一遍扫描之后就可以确定start位置，之后再追加一个N次扫描，因此至多为2N的时间。代码如下：

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int m = gas.size();
        int count = 0, sum = 0, pos = 0, start = 0;
        vector<int> left;
        for (int i = 0; i < m; i++)
            left.push_back(gas[i] - cost[i]);
            
        for(int i = 0; i < m; i++)
            sum += left[i];
        if (sum < 0)
            return -1;
        
        sum = 0;
        while (count < m){          //利用count来记录走过的站点数
            sum += left[pos];
            pos = ++pos % m;        //取模来把超出范围的pos值拉回到0， m % m = 0 既第m+1 个pos又置0
            if (sum < 0){
                sum = 0;
                count = 0;
                start = pos;
            }
            else{
                count++;
            }
        }
        return start;
    }
};

LeetCode :: Gas Station