[LeetCode]Gas Station@Python

最新推荐文章于 2025-09-06 22:26:52 发布

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最新推荐文章于 2025-09-06 22:26:52 发布

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分类专栏： leetcode greedy 文章标签： leetcode python

本文链接：https://blog.youkuaiyun.com/byr_wy/article/details/91488191

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本文探讨了一种特殊的情境：在环形路线上，有N个加油站，每个加油站提供一定数量的汽油。驾驶一辆拥有无限油箱容量的汽车，从任意加油站开始旅程，每行驶到下一个加油站需要消耗一定的汽油。文章提出了一种解决方案，用以确定能否从某个加油站出发并返回原点，同时保证油箱中的汽油不会耗尽。若存在这样的起点，则返回其索引；反之，则返回-1。

Gas Station

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 in the clockwise direction, otherwise return -1.
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.

Example

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.

Solution

# unique solution decides that we don't need to calculate the max value
class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """
        :type gas: List[int]
        :type cost: List[int]
        :rtype: int
        """
        if sum(gas)<sum(cost):
            return -1
        n, s, r = len(gas), 0, 0
        for i in range(n):
            if gas[i]+r<cost[i]:
                s, r = i+1, 0
            else:
                r+= gas[i]-cost[i]
        return s