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.

思路： 设置以当前的 sum  和一个 total  的 sum ， 如果当前的sum < 0 ，则从这个开始重新开始。 注意 最后检查是否可以到达一圈， 也就是 total >= 0 . 

易错点： 最后返回 index + 1， 初始化为index = -1；

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