# 134. Gas Station (1)

[Link](https://leetcode.com/problems/gas-station/)

**Description**

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 in the clockwise direction, otherwise return -1.

**Note:**

* If there exists a solution, it is guaranteed to be unique.
* Both input arrays are non-empty and have the same length.
* Each element in the input arrays is a non-negative integer.

**Example 1:**

```
Input: 
gas  = [1,2,3,4,5]
cost = [3,4,5,1,2]

Output: 3

Explanation:
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.
```

**Example 2:**

```
Input: 
gas  = [2,3,4]
cost = [3,4,3]

Output: -1

Explanation:
You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
Travel to station 0. Your tank = 4 - 3 + 2 = 3
Travel to station 1. Your tank = 3 - 3 + 3 = 3
You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
Therefore, you can't travel around the circuit once no matter where you start.
```

**Solution**

常常用一定可以找出一個解的迴圈過濾掉一定不可能的 ，但找到的不一定完全符合，只能確保滿足部份條件，若有解就是它了。

最後仍要在進行檢查是否有解。

```
class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int sumGas = 0;
        int sumCost = 0;
        int start = 0;
        int tank = 0;
        //如果A->B->C->D->E, 若A不能到D，則B,C一定都不能到D
        //然後D一定是gas比cost少，所以也不可能當出發點
        //所以只有可能是E以後的可能是出發點(本次無法到達的下個點)
        for (int i = 0; i < gas.length; i++) {
            sumGas += gas[i];
            sumCost += cost[i];
            tank += gas[i] - cost[i];
            if (tank < 0) {
                //假設下一個當start去試
                start = i + 1;
                tank = 0;
            }
        }
        
        
        //最後還需要針對不可能找到的case 例外處理
        if (sumGas < sumCost) {
            return -1;
        } else {
            return start;
        }
    }
}
```
