算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## 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

首先想到的是$$O(N^2)$$的解法,对每个点进行模拟。

`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

{% if book.java %}
{% codesnippet "./code/gas-station."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}

{% if book.cpp %}
```cpp
// Gas Station
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
int canCompleteCircuit(vector &gas, vector &cost) {
int total = 0;
int j = -1;
for (int i = 0, sum = 0; i < gas.size(); ++i) {
sum += gas[i] - cost[i];
total += gas[i] - cost[i];
if (sum < 0) {
j = i;
sum = 0;
}
}
return total >= 0 ? j + 1 : -1;
}
};
```
{% endif %}