[leetcode] 134. Gas Station

原创已于 2024-05-07 22:57:49 修改 · 168 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #算法 #职场和发展

于 2019-02-12 23:35:53 首次发布

leetcode题解同时被 3 个专栏收录

903 篇文章

订阅专栏

python

556 篇文章

订阅专栏

C++

441 篇文章

订阅专栏

本文探讨了一个经典的加油站路径问题，给出了一种高效的算法解决方案。通过分析加油站的油量和行驶成本，算法能找出从哪个加油站出发可以顺时针完成一圈旅行。文章提供了两种不同的算法思路，并附带了详细的代码实现。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Description

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.

分析一

题目的意思是：给你一堆加油站，和去每个加油站的消耗，要求找到从某个位置开始能够顺时针遍历一圈。

记录最后一个加起来小于零的索引，然后返回这个索引+1就是答案了。

代码一

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int total=0;
        int sum=0;
        int start=0;
        for(int i=0;i<gas.size();i++){
            total+=gas[i]-cost[i];
            sum+=gas[i]-cost[i];
            if(sum<0){
                start=i+1;
                sum=0;
            }
        }
        return total>=0 ? start:-1;
    }
};

分析二

从start出发，如果油量够，可以一直向后走end++;油量不够的时候，
start向后退，最终start==end的时候，如果有解一定是当前start所在位置。

代码二

class Solution {
public:
    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {
        int start=gas.size()-1;
        int end=0;
        int sum=gas[start]-cost[start];
        while(start>=end){
            if(sum>=0){
                sum+=gas[end]-cost[end];
                end++;                
            }else{
                start--;
                sum+=gas[start]-cost[start];   
            }
        }
        return sum>=0 ? start:-1;
    }
};

Python实现

如果x到达不了y+1，那么x-y之间的点也不可能到达y+1，因为中间任何一点的油都是拥有前面的余量的，所以下次遍历直接从y+1开始

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