题目：
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:
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.length == cost.length
1 <= n <= 105
0 <= gas[i], cost[i] <= 104
解题思路：
这道题目中，我们已知每个油站可以获得的油量，以及前往下一个油站的油耗，我们需要判断是否有一个油站可以作为起始点，让车子有足够的油耗遍历每个油站，跑完一整圈。
如果所有油站的油量总和要小于油耗的总和，车子一定跑不完一圈，返回 -1。
答案是唯一的，所以有且只有一个油站可以让车子跑完一圈。
首先，检查是否所有油站的油量总和要大于油耗的总和。
使用贪婪算法找到这个能跑完一圈的起始油站。通过记录现有的油量 current gas，如果一个油站的现有油量小于0，那么我们一定不可能继续跑完一整圈。所以我们把下一个油站假设为起始点，重设 current gas为0，再作检查。
class Solution(object):def canCompleteCircuit(self, gas, cost):""":type gas: List[int]:type cost: List[int]:rtype: int"""total_gas = 0current_gas = 0start_position = 0for i in range(len(gas)):total_gas += gas[i] - cost[i]current_gas += gas[i] - cost[i]if current_gas < 0:start_position = i + 1 # 设置下一个油站为我们想要找的起始点current_gas = 0if total_gas < 0:return -1else:return start_position
time complexity为O(n)