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, otherwise return -1

题意：一个环路中有N个加油站，每个加油站有gap[i]的汽油，从加油站i到i+1耗油为cost[i]，问一个油箱为无限大的汽车，能不能走完全程，能的话起点在哪

思路：如果环路中gas的总量比cost的总量多，那么肯定有一种方法可以让汽车跑完全程，首先什么情况下汽车不能跑完全程？比如从i出发到i+n时候，邮箱变为负值了，那么就可以把从i到i+n的节点全部滤掉

做法和求最大子列和一样

 1 class Solution(object):
 2     def canCompleteCircuit(self, gas, cost):
 3         if not len(gas) or not len(cost) or sum(gas)<sum(cost):
 4             return -1
 5         balance,p= 0,0
 6         for i in range(len(gas)):
 7             balance += gas[i]-cost[i]
 8             if balance < 0:
 9                 balance = 0
10                 p = i+1
11         return p

[leetcode greedy]134. Gas Station