Gas Station leetcode

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.

用暴力求解方式 两个变量sum和total存储每一站邮箱的剩余量gas[i] - cost[i]
如果sum[i]<0 说明无法走到i+1点 从这个地方开始走并不合适 把sum重置为0 

而且不仅这个起点不能作为起点 自起点到sum < 0 的i点的点都不能作为起点 假设能从k (i =>>  k=>> j)走，那么i..j < 0，若k..j >=0，说明i..k – 1更是<0，那从k处sum就小于0了，根本轮不到j。那么如果起点都不可以 中间的点就都不可以

因为如果当前次不行就要从i+1开始走 走到i= lengh时候toal记录了从头到尾的gas-cost 所以如果total为+ 那么久可以走完 但是total为- 就不可以
时间O(n) 空间 O(1)

public class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        if (gas == null || cost == null || gas.length == 0 || cost.length == 0) {
            return -1;
        }
        int sum = 0;
        int total = 0;
        int index = 0;
        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 + 1;
            }
        }
        if (total < 0) {
            return -1;
        } else {
            return index;
        }
    }
}