Question

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 ⑴.
在1個圈里面有N個汽油站，i站的汽油有gas[i]汽油。現(xiàn)在有1輛無窮容量油箱的車，它從i站開到（i+1）需要耗費cost[i]汽油。如果這輛車可以走完這個圈，那末返回這個車的出發(fā)點，否者返回⑴.

Algorithm

• 如果從A到不了B，那末A-B之間的任何1站都到不了B（B是從A開始第1個不能到達的站）
• 如果各個站的油量總和超過需要消耗油量總和，則1定有解

Accepted Code

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