【LeetCode】134. Gas Station

最新推荐文章于 2024-05-10 11:05:44 发布

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最新推荐文章于 2024-05-10 11:05:44 发布

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原文链接：http://www.cnblogs.com/ganganloveu/p/4117531.html

本文探讨了一种特定的算法问题——环路加油站问题。该问题要求寻找一条从某个加油站出发可以环绕整个环路的路径。文章提供了一种O(n)复杂度的解决方案，并通过实例解释了如何判断是否有可行的起点。

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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.

Note:
The solution is guaranteed to be unique.

思路：顺序扫一圈。如果总油量小于总消耗，说明起点错误，则将起点往后移位。

复杂度O(n)。

几个地方需要注意：

1、如果有解，则终点绕一圈之后回到起点

2、最多绕一圈即可判断是有解还是无解，起点重复代表无解

3、终点必须在起点之后

class Solution {
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        if(gas.empty())
            return -1;
            
        int begin = 0;
        int end = begin;
        int n = gas.size();
        int totalgas = gas[begin];
        int totalcost = cost[begin];
        while(true)
        {
            while(totalgas >= totalcost)
            {
                end = (end+1)%n;
                if(end == begin)
                    return begin;
                totalgas += gas[end];
                totalcost += cost[end];
            }
            totalgas -= gas[begin];
            totalcost -= cost[begin];
            begin = (begin+1)%n;
            end = max(begin, end);  //end >= begin
            if(begin == end)
            {// restart
                totalgas = gas[begin];
                totalcost = cost[begin];
            }    
            if(begin == 0)
                return -1;
        }
    }
};