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.

这道转圈加油问题不算很难，只要相同其中的原理就很简单。我们首先要知道能走完整个环的前提是gas的总量要大于cost的总量，这样才会有起点的存在。假设开始设置起点start = 0, 并从这里出发，如果当前的gas值大于cost值，就可以继续前进，此时到下一个站点，剩余的gas加上当前的gas再减去cost，看是否大于0，若大于0，则继续前进。当到达某一站点时，若这个值小于0了，则说明从起点到这个点中间的任何一个点都不能作为起点，则把起点设为下一个点，继续遍历。当遍历完整个环时，当前保存的起点即为所求。代码如下：

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

[LeetCode] Gas Station 加油站问题