## Gas Station

### 描述

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.

### 分析

O(N)的解法是，设置两个变量，sum判断当前的指针的有效性；total则判断整个数组是否有解，有就返回通过sum得到的下标，没有则返回-1。

### 代码

// Gas Station
// 时间复杂度O(n)，空间复杂度O(1)
public class Solution {
public int canCompleteCircuit(int[] gas, int[] cost) {
int total = 0;
int j = -1;
for (int i = 0, sum = 0; i < gas.length; ++i) {
sum += gas[i] - cost[i];
total += gas[i] - cost[i];
if (sum < 0) {
j = i;
sum = 0;
}
}
};