134. Gas Station Add to List

There are N gas stations along a circular route, where the amount of gas at stationi is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from stationi 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.

本题采用贪心算法，若从i点无法到达k点，那么从i~k的任意一点都无法到达k。同时若gas的总和大于cost的总和，则说明必然有解。

所以核心就是寻找所谓的start，设定tank变量存储当前gas和cost的差，当tank小于0，说明存在上述i~k的情况，便将start跳到tank之后。一轮检索之后，比较总的gas与cost，若大于等于0，则返回cost；小于0则意味着无法转一圈，则返回-1.

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int start(0),total(0),tank(0);
        for(int i=0;i<gas.size();i++) if((tank=tank+gas[i]-cost[i])<0) {start=i+1;total+=tank;tank=0;}
        return (total+tank<0)? -1:start;
    }
};