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.

解法：這道題的重點在於理解：當車從i跑不到j的時候,那麼從i到j的中間任何一點也都跑不到j，因為能跑到某個站的前提是現有汽油數大於等於消耗數，因此這道題的思路就是，從0開始跑，直到跑不到i,那麼就從i重新跑，直到從某個位置j能跑到0，這時候我們再檢查從0能不能跑到j,如果可以就返回j,不可以則返回-1,程式碼如下：

    public int canCompleteCircuit(int[] gas, int[] cost) {
        int gass = gas[0], start = 0
 
;
        for (int i = 0; i < gas.length - 1; i++) {
            gass -= cost[i];
            if (gass >= 0) {
                gass += gas[i+1];
            } else {
                start = i + 1;
                gass = gas[i + 1];
            }
        }
        gass -= cost[gas.length - 1];
        if
 
 (gass < 0) {
            return -1;
        }
        gass += gas[0];
        for (int i = 0; i < start; i++) {
            gass -= cost[i];
            if (gass >= 0) {
                gass += gas[i+1];
            } else {
                return -1;
            }
        }
        return start;
    }