Gas Station

原题：
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：
先用gas[i]-cost[i]， 然后求取连续那个元素的序列和都大于0的串，题目没有要求输出所有合适的，默认找到合适的start position就返回。可以AC的C++代码如下：

    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {        if(gas.size() < 1 || cost.size() < 1)            return -1;        if(gas.size() != cost.size())            return -1;        int n = gas.size();        int sum = 0, index = -1, cnt = 0;        for(int i=0; i<2*n-1; i++){            sum += gas[i%n] - cost[i%n];            if(sum >= 0){                if(index < 0)                   index = i % n;                cnt ++;            }else{                sum = 0;                index = -1;                cnt = 0;            }            if(cnt == n)               break;        }        if(cnt < n){            index = -1;        }        return index;    }