# Gas Station

There are `n` gas stations along a circular route, where the amount of gas at the `ith` station is `gas[i]`.

You have a car with an unlimited gas tank and it costs `cost[i]` of gas to travel from the `ith` station to its next `(i + 1)th` station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays `gas` and `cost`, return *the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return* `-1`. If there exists a solution, it is **guaranteed** to be **unique**

**Example 1:**

<pre><code><strong>Input: gas = [1,2,3,4,5], cost = [3,4,5,1,2]
</strong><strong>Output: 3
</strong><strong>Explanation:
</strong>Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 4. Your tank = 4 - 1 + 5 = 8
Travel to station 0. Your tank = 8 - 2 + 1 = 7
Travel to station 1. Your tank = 7 - 3 + 2 = 6
Travel to station 2. Your tank = 6 - 4 + 3 = 5
Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
Therefore, return 3 as the starting index.
</code></pre>

***My Solutions:***

可以先遍历一次看看是否有解，如果sum of gas < sum of cost 一定无解。

因为知道一定有解，所以从index=0开始，遍历到最后，每一次如果汽车tank的油量小于0，说明当前的index是不能用的，此时让index+1，清除tank。

也可以优化一下，不需要先遍历一遍判断有没有解，而是多一个在遍历时计算tank total值。最后检查是不是<0。

```
public int canCompleteCircuit(int[] gas, int[] cost) {
    if (gas == null || cost == null || gas.length == 0 || cost.length == 0) return -1;
    int gasSum = 0, costSum = 0;
    for (int i = 0; i < gas.length; i++) {
        gasSum += gas[i];
        costSum += cost[i];
    }
    if (gasSum < costSum) return -1;

    // 也可以多一个total值 total = 0;
    int tank = 0, index = 0;
    for (int i = 0; i < gas.length; i++) {
        tank += gas[i] - cost[i];
        if (tank < 0) {
            index = i + 1;
            tank = 0;
        }
        // 在这里算总值 total += gas[i] - cost[i];
    }
    // 在这里返回 if (total < 0) return -1;
    return index;
}
```
