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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: 3Explanation:Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4Travel to station 4. Your tank = 4 - 1 + 5 = 8Travel to station 0. Your tank = 8 - 2 + 1 = 7Travel to station 1. Your tank = 7 - 3 + 2 = 6Travel to station 2. Your tank = 6 - 4 + 3 = 5Travel 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.

gas代表加油站有多少油，cost代表去下一站需要用多少油，最后返回在那一站出发可以走完全程（回到出发点）。

首先可以用sum来判断是否能走完全程，如果sum(cost)>sum(gas)，这种情况无论从哪里开始都不可能实现走完全程。

当sum(cost)<=sum(gas)时，一定存在某个起点，可以走完全程。

所以用一个for循环去寻找即可。

res表示油箱中剩余油量，start表示起点。

当某个时间res为负的时候，代表之前的点都不能作为起点，

这时start设为i+1，从新起点出发，res重置为0。

class Solution:    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:        if sum(gas)

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