Gas Station

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.

思路：

1.暴力。O(n^2)

2.如果总的gas-cost小于0，那么返回-1；如果大于0，遍历每个gas station,如果当前索引走不到，那么重新标记下一个索引，重新计数知道循环结束，返回标记索引。

sum判断当前的指针的有效性，total则判断整个数组是否有解，如果有则通过sum的到下标，没有返回-1。

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