加油站

中英题面

在一条环路上有N个加油站，其中第i个加油站有汽油gas[i]升。

There areNgas stations along a circular route, where the amount of gas at stationiisgas[i].

你有一辆油箱容量无限的的汽车，从第i个加油站开往第i+1个加油站需要消耗汽油cost[i]升。你从其中的一个加油站出发，开始时油箱为空。

You have a car with an unlimited gas tank and it costscost[i]of gas to travel from stationito its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

如果你可以绕环路行驶一周，则返回出发时加油站的编号，否则返回 -1。

Return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1.

说明:

• • 如果题目有解，该答案即为唯一答案。
  • 输入数组均为非空数组，且长度相同。
  • 输入数组中的元素均为非负数。

Note:

• • If there exists asolution, it is guaranteed to be unique.
  • Both input arrays are non-empty and have the same length.
  • Each element in the input arrays is a non-negative integer.

示例1:

　　输入: 
　　gas  = [1,2,3,4,5]
　　cost = [3,4,5,1,2]

　　输出: 3

　　解释:
　　从 3 号加油站(索引为 3 处)出发，可获得 4 升汽油。此时油箱有 = 0 + 4 = 4 升汽油
　　开往 4 号加油站，此时油箱有 4 - 1 + 5 = 8 升汽油
　　开往 0 号加油站，此时油箱有 8 - 2 + 1 = 7 升汽油
　　开往 1 号加油站，此时油箱有 7 - 3 + 2 = 6 升汽油
　　开往 2 号加油站，此时油箱有 6 - 4 + 3 = 5 升汽油
　　开往 3 号加油站，你需要消耗 5 升汽油，正好足够你返回到 3 号加油站。
　　因此，3 可为起始索引。

Example 1:

　　Input: 
　　gas  = [1,2,3,4,5]
　　cost = [3,4,5,1,2]
　　
　　Output: 3
　　
　　Explanation:
　　Start at station 3 (index 3) and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
　　Travel to station 4. Your tank = 4 - 1 + 5 = 8
　　Travel to station 0. Your tank = 8 - 2 + 1 = 7
　　Travel to station 1. Your tank = 7 - 3 + 2 = 6
　　Travel to station 2. Your tank = 6 - 4 + 3 = 5
　　Travel to station 3. The cost is 5. Your gas is just enough to travel back to station 3.
　　Therefore, return 3 as the starting index.

示例 2:

　　输入: 
　　gas  = [2,3,4]
　　cost = [3,4,3]

　　输出: -1

　　解释:
　　你不能从 0 号或 1 号加油站出发，因为没有足够的汽油可以让你行驶到下一个加油站。
　　我们从 2 号加油站出发，可以获得 4 升汽油。 此时油箱有 = 0 + 4 = 4 升汽油
　　开往 0 号加油站，此时油箱有 4 - 3 + 2 = 3 升汽油
　　开往 1 号加油站，此时油箱有 3 - 3 + 3 = 3 升汽油
　　你无法返回 2 号加油站，因为返程需要消耗 4 升汽油，但是你的油箱只有 3 升汽油。
　　因此，无论怎样，你都不可能绕环路行驶一周。

Example 2:

　　Input: 
　　gas  = [2,3,4]
　　cost = [3,4,3]

　　Output: -1

　　Explanation:
　　You can't start at station 0 or 1, as there is not enough gas to travel to the next station.
　　Let's start at station 2 and fill up with 4 unit of gas. Your tank = 0 + 4 = 4
　　Travel to station 0. Your tank = 4 - 3 + 2 = 3
　　Travel to station 1. Your tank = 3 - 3 + 3 = 3
　　You cannot travel back to station 2, as it requires 4 unit of gas but you only have 3.
　　Therefore, you can't travel around the circuit once no matter where you start. <br /><br /><br /><br />

算法

设f[i] = gas[i] – cost[i]。

无解当且仅当sum{f[i]} < 0。

通过队列维护从每个加油站出发的f[i]的前缀和s[i]，当s[i]恒为0时，则i为一个可行的起点。

时间复杂度：

O(N)

空间复杂度：

O(1)

代码

1 class Solution:
 2     def canCompleteCircuit(self, gas, cost):
 3         """
 4         :type gas: List[int]
 5         :type cost: List[int]
 6         :rtype: int
 7         """
 8         n = len(gas)
 9         for i in range(n):
10             gas[i] -= cost[i]
11         if (sum(gas) < 0):
12             return -1
13         i = j = now = 0
14         while (True):
15             now += gas[i]
16             i = (i + 1) % n
17             if (i == j):
18                 return j
19             while (now < 0):
20                 now -= gas[j]
21                 j = (j + 1) % n