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 are N gas stations along a circular route, where the amount of gas at station i is gas[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 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.
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 a solution, 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.
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)
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
