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：

[code]    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;
    }