加油站绕圈-数组倍长

在一个环上有n个加油站，已知第i个加油站可以提供ai的油量，从第i个加油站到第i+1个加油站需要消耗bi的油量。（当i为n时，则表示n到1的油量）

问：是否能从某个加油站出发，绕环一圈回到这个加油站？（当途中到达某个加油站时的油量小于0则不可）。

 

    public static void main(String[] args) {
        int[] gas = {4,2, 3, 1};
        int[] cost = {1,3, 1, 2};
        System.out.println(canCompleteCircuit2(gas, cost));
    }

    /**
     * @param gas=[4,2,3,1]
     * @param cost=[1,3,1,2]
     * @return
     */
    public static int canCompleteCircuit2(int[] gas, int[] cost) {
        int[] gasDouble = new int[gas.length * 2];
        for (int i = 0; i < gasDouble.length; i++) {
            if (i < gas.length) {
                gasDouble[i] = gas[i];
            } else {
                gasDouble[i] = gas[i - gas.length];
            }
        }
        int[] costDouble = new int[cost.length * 2];
        for (int i = 0; i < costDouble.length; i++) {
            if (i < cost.length) {
                costDouble[i] = cost[i];
            } else {
                costDouble[i] = cost[i - cost.length];
            }
        }
        for (int i = 0; i < gasDouble.length; i++) {
            boolean flag = true;
            int left = 0;
            for (int j = i; j < i + gas.length; j++) {
                left += gasDouble[j];
                if (left < costDouble[i]) {
                    flag = false;
                    break;
                }
                left -= costDouble[i];
            }
            if (flag) {
                return i;
            }
        }
        return -1;
    }

算法参考：https://www.cnblogs.com/andyqsmart/p/5460020.html