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  • leetcode------Gas Station

    标题: Gas Station
    通过率: 25.7%
    难度: 中等

    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.

    Note:
    The solution is guaranteed to be unique.

    本题看半天不明白什么意思,网上有一大堆的解法各种动态规划什么的。我感觉都不靠谱,下面我们分析:

    先分析暴力破解,就是从每一个点出发计算一次,看看最后是否成立。那么时间复杂度是O(n^2)

    其实一遍就能知道,

    如从一个点p出发到k时油量小于0,下一次试探肯定是从k开始,

    若从k开始刚好可以到数组的最后,那么不用不去测试从数组最后返回前面,

    因为若能循环一次那么gas-cost总和一定大于0.所以只用确定有一个点出发可以走到最后,然后看total是否大于0,

    具体看代码:

     1 class Solution:
     2     # @param gas, a list of integers
     3     # @param cost, a list of integers
     4     # @return an integer
     5     def canCompleteCircuit(self, gas, cost):
     6         if len(gas)==0 or len(cost)==0 or len(gas)!=len(cost):return -1
     7         start,total,sum=0,0,0
     8         for i in range(len(gas)):
     9             total+=(gas[i]-cost[i])
    10             if sum<0:
    11                 sum=gas[i]-cost[i]
    12                 start=i
    13             else :sum+=(gas[i]-cost[i])
    14         if total<0:return -1
    15         else: return start
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  • 原文地址:https://www.cnblogs.com/pkuYang/p/4430243.html
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