leetcode134 -- Gas Station (被面过)

leetcode134 -- Gas Station (被面过)

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 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 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.

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.

这个题的意思是有一些加油站刚好弄成了一个环,问能否找到一个出发点,从这个点出发能够环游一遍.

解答

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        total = 0
        summ = 0
        start = 0
        
        for i in range(len(gas)):
            total += gas[i]-cost[i]
            summ += gas[i]-cost[i]
            
            if summ<0:
                summ = 0
                start = i+1
                continue
                
        if total>=0:
            return start
        return -1
        

虽然题目中说的是一个环,但是其实可以不必考虑环,直接就是list就行,这里的total是到达第(i+1)个加油站时的存油量,初始化为0,summ 是到达下一个站加油站时的存油量,如果它为负的话就把它清0,因为这时候它不了下一个加油站,需要重设起点,并把起点更新为下一个加油站. 最终total如果非负的话,说明就可以走完,否则的话就走不完。

这里可能会有question是当以中间的某个为起点时,走到最后的时候并没有走完所有的啊,是如何判断能够走完的呢,关键就是用了total和summ这两个变量。

比如说有10个加油站,现在前面的0到4都不能作为起点,5能够作为起点,并且从5能走到最后,那么如果最后total非负的话,那它一定还能够走回到5。理由是当5走到9时,此时total非负,summ也非负(否则5不会是起点),说明可以从9走到0,然后这时候如果它接着从9走到0的话,那么这时候 从5出发走到9再走到0的total=summ>=0(因为中间没有summ<0的,summ没有reset),因为前面的0到4都不是起点,所以它们几个的加起来的total是负的,但是总的total是正的,所以从5走到9再走到0之后仍然可以往前走完他们的每一个。因此是可以走完的。

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