题目描述
环形路上有n个加油站，第i个加油站的汽油量是gas[i].
你有一辆车，车的油箱可以无限装汽油。从加油站i走到下一个加油站（i+1）花费的油量是cost[i]，你从一个加油站出发，刚开始的时候油箱里面没有汽油。
求从哪个加油站出发可以在环形路上走一圈。返回加油站的下标，如果没有答案的话返回-1。
注意：
答案保证唯一。
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.
Note:
The solution is guaranteed to be unique.
解题思路
从start出发，油量够就一直向后走end++，
不够的话，start向后退，这样就能够在油量不够的时候往后遍历新的出发点，最后start == end 的时候，如果有解就是start的位置。
class Solution { public: int canCompleteCircuit(vector<int> &gas, vector<int> &cost) { int start = gas.size() - 1; int end = 0; int sum = gas[start] - cost[start]; while( start > end){ if(sum >= 0){ sum += gas[end] - cost[end]; end++; } else{ start--; sum += gas[start] - cost[start]; } } return sum >= 0 ? start : -1; } };
