zoukankan      html  css  js  c++  java
  • [leetcode greedy]134. 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, otherwise return -1

    题意:一个环路中有N个加油站,每个加油站有gap[i]的汽油,从加油站i到i+1耗油为cost[i],问一个油箱为无限大的汽车,能不能走完全程,能的话起点在哪

    思路:如果环路中gas的总量比cost的总量多,那么肯定有一种方法可以让汽车跑完全程,首先什么情况下汽车不能跑完全程?比如从i出发到i+n时候,邮箱变为负值了,那么就可以把从i到i+n的节点全部滤掉

    做法和求最大子列和一样

     1 class Solution(object):
     2     def canCompleteCircuit(self, gas, cost):
     3         if not len(gas) or not len(cost) or sum(gas)<sum(cost):
     4             return -1
     5         balance,p= 0,0
     6         for i in range(len(gas)):
     7             balance += gas[i]-cost[i]
     8             if balance < 0:
     9                 balance = 0
    10                 p = i+1
    11         return p
  • 相关阅读:
    Bzoj3339 Rmq Problem
    Bzoj3509 [CodeChef] COUNTARI
    浅析python日志重复输出问题
    mysql练习题
    python学习之思维导图
    python面向对象编程练习
    Python常见下划线
    内置方法
    类的绑定方法与非绑定方法
    封装
  • 原文地址:https://www.cnblogs.com/fcyworld/p/6533665.html
Copyright © 2011-2022 走看看