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.
1 如果总的gas - cost小于零的话，那么没有解返回-1。（总gas> 总cost 是有解的充分便要条件，要是解这道题的关键）
2 如果前面所有的gas - cost加起来小于零，那么前面所有的点都不能作为出发点（存在gas-cost<0，剩下的必然存在gas-cost>0）。
3 选择最佳出发点后，判断从这点出发能否环行一圈。
class Solution {
public:
int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
int len = gas.size();
int res = -1;
int res_gas = 0;
int start_station = 0;
int total_res = 0;
for(int i =0 ; i < len; i++)
{
res_gas += (gas[i] - cost[i]) ;
total_res += (gas[i] - cost[i]) ;
if(res_gas < 0)
{
start_station = i+1;
res_gas = 0;
}
}
if(total_res >= 0)
{
res = start_station;
}
return res;
}
};