leetCode No.134 Gas Station

题目

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.

标签:Greedy

题意

有n个加油站形成了一个环。每个加油站都有gas[i]数量的汽油,汽车加油站i移动到加油站i+1需要汽油cost[i]。现给定gas和cost数组,求是否存在一个加油站的位置,可以让一辆一开始没有油的汽车走完所有加油站。

解题思路

汽车到达一个站会获得gas[i]数量的汽油,前往下一站(i+1站)会消耗cost[i]的汽油,对车的汽油数一直记录,若当前汽车的油量已经为负数说明已经不能继续进行,这时将累计的汽油数清零,从下一个车站开始记录。需要注意的是这样的算法不能判断问题是否有可行解,这样就需要在最开始的时候定义一个total,每次汽车前进都用total = gas[i] - cost[i]来累计total的量。当最后total > 0则有可行解,反之则无。

代码

class Solution(object):
    def canCompleteCircuit(self, gas, cost):
        """ :type gas: List[int] :type cost: List[int] :rtype: int """
        total = 0
        start = 0
        m_sum = 0
        for i in range(len(gas)):
            total += gas[i] - cost[i]
            if m_sum < 0:
                start = i
                m_sum = gas[i] - cost[i]
            else:
                m_sum += gas[i] - cost[i]
        if total < 0:
            return -1
        else:
            return start

相关链接

源代码(github)
原题

你可能感兴趣的:(LeetCode)