leetcode134. Gas Station-CSDN博客

本文探讨了一个经典的算法问题：环路加油站问题。问题中有一系列加油站分布在环形路线上，每站提供一定数量的汽油，并且从一站到下一站需要消耗一定量的汽油。任务是找到一个起点，使得车辆可以从该点出发并绕环形路线一周后回到起点。文章提供了两种解决方案，一种是直观的模拟方法，复杂度为O(N^2)，另一种则是高效的线性时间复杂度算法，即O(N)。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

medium程度题

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.

Note:
The solution is guaranteed to be unique.

我的想法是对每个点进行模拟，复杂度O(N * N)

AC解

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) 
    {
        int size = gas.size();
        vector<int> gap(size,0);
        for (int i = 0; i < size; i++)
            gap[i] = gas[i] - cost[i];
        
        int i = 0,j = 0;
        int sum = 0;
        for (i = 0; i < size; i++)
        {
            sum = 0;
            for (j = i; j < size; j++)
            {
                sum += gap[j];
                if (sum < 0)  break;
            }
            //如果到最后一个元素了
            if (j == size)
            {	//如果是从第一个站开始的，则行驶完了一圈
                if (i == 0)
                    break;
                    
                int m = 0;
                for (m = 0; m < i; m++)
                {
                    sum += gap[m];
                    if (sum < 0)  break;
                }
                //行驶完了一圈
                if (m == i)  break;
            }
            
        }
        
        if (i == size)
            return -1;
        return sum >= 0 ? i : -1;
    }
};

别人的高效解法,复杂度O(N)

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) 
    {
        int total = 0;//一圈计算完后total >= 0 则有解
        int result = -1;
        int size = gas.size();
        for (int i = 0,sum = 0; i < size; i++)
        {
            sum += gas[i] - cost[i];
            total += gas[i] - cost[i];
            
            if (sum < 0)
            {
                result = i;
                sum = 0;
            }
        }
        
        return total >= 0 ? result + 1 : -1;
    }
};