134.Gas Station
===

###### tags: `Medium`,`Array`,`Greedy`

[134. Gas Station](https://leetcode.com/problems/gas-station/)

### 題目描述

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

You have a car with an unlimited gas tank and it costs `cost[i]` of gas to travel from the i^th^ 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**

### 範例

**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.

**Constraints**:

* `n` == `gas.lengt`h == `cost.length`
* 1 <= `n` <= 10^5^
* 0 <= `gas[i]`, `cost[i]` <= 10^4^

### 解答

#### Python

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        start = 0
        total = 0
        remaining = 0
        for i in range(len(gas)):
            total += gas[i] - cost[i]
            remaining += gas[i] - cost[i]
            if remaining < 0:
                start = i + 1
                remaining = 0
        return start if total >= 0 else -1

> [name=Yen-Chi Chen][time=Sun, Jan 7, 2023]

#### Javascript

function canCompleteCircuit(gas, cost) {
  let total = 0;
  let tank = 0;
  let start = 0;

  for (let i = 0; i < gas.length; i++) {
    const leftover = gas[i] - cost[i];
    tank += leftover;
    total += leftover; // 加總所有的剩餘油量，如果小於0，代表沒辦法走完一圈
    
    // 如果目前的油量小於0，代表從目前的站點開始，無法走完一圈
    if (tank < 0) {
      start = i + 1;
      tank = 0;
    }
  }

  if (total < 0) return -1;

  return start;
}

> 月底壓線補一題QQ
> [name=Marsgoat][time=Jan 31, 2023]

### Reference

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