# 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.

这题的意思就是求出从哪一个油站开始,能走完整个里程,并且这个结果是唯一的。

首先我们可以得到所有油站的油量totalGas,以及总里程需要消耗的油量totalCost,如果totalCost大于totalGas,那么铁定不能够走完整个里程。

如果totalGas大于totalCost了,那么就能走完整个里程了,假设现在我们到达了第i个油站,这时候还剩余的油量为sum,如果 sum + gas[i] - cost[i]小于0,我们无法走到下一个油站,所以起点一定不在第i个以及之前的油站里面(都铁定走不到第i + 1号油站),起点只能在i + 1后者后面。

代码如下:

```c++
class Solution {
public:
int canCompleteCircuit(vector &gas, vector &cost) {
int sum = 0;
int total = 0;
int k = 0;
for(int i = 0; i < (int)gas.size(); i++) {
sum += gas[i] - cost[i];
//小于0就只可能在i + 1或者之后了
if(sum < 0) {
k = i + 1;
sum = 0;
}
total += gas[i] - cost[i];
}

if(total < 0) {
return -1;
} else {
return k;
}
}
};
```