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:这道题可以基于最大字数列和的思想来解题,每次循环累加和存油量与耗油量只差分别记录两次,如果这次循环的差和小于零,则start记录这个点的下一个,并且将上述的其中一个变量置零,进行下次循环。如果循环下来,上述差额的累加和小于零的话,说明无法完成旅程。反之,返回start,记为开始旅程的起点。

class Solution {

public:

    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {

        if(gas.size()==0)

            return 0;

        int sum=0;

        int diff_sum=0;

        int result=0;

        for(int i=0;i<gas.size();i++)

        {

            sum+=gas[i]-cost[i];

            diff_sum+=gas[i]-cost[i];

            if(sum<0)

            {

                result=(i+1)%gas.size();

                sum=0;

            }

        }

        if(diff_sum<0)

            return -1;

        return result;

    }

};

 

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