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.

我感觉这是一道比较trick的题目，用贪心算法，解决的关键在于：1，如果从A点出发中途经过几个加油站之后不能到达X点，那么A到X之间任何加油站都不能到达X点。2，如果总的油量大于路上需要消耗的油量，那么一定存在一个加油站，从它开始出发可以走完所有的加油站。知道了上面的两点就可以解决这道题目了，用remain来记录当前的油量，用tatal来记录总的油量，当remain小于0的时候，就说明之前出发的加油站不可以，要从下一个加油站出发，用start来记录从哪个加油站出发。代码如下：

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