[LeetCode] 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.

Solution:

思路1：通过两层循环，依次从某个点出发并测试是否能够运行一圈。时间复杂度为O（n2），不满足要求。

思路2：首先确认gas总和大于cost，因此判断能够绕圈。接下来寻找起始位置，我们可以借鉴归并排序的思路，如果某一段路gas>cost，则这段路剩余的油量可以支撑其他路段。因此问题变化为找到某个节点，在它之前的路段剩余油量为负，而从它开始到整个队列结束剩余油量为正（正油量可以不足前面路段的不足油量）。时间可以在O（n）完成。(分析 From CSDN)

class Solution {

public:

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

        int length = gas.size();

        int *sum = new int[length];

        for(int i = 0;i < length;i++)

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

        

        int remain = 0;

        bool flag;

        for(int i = 0;i < length;i++)

        {

            if(sum[i] >= 0)

            {

                //start from i

                flag = true;

                remain = 0;

                for(int j = i;j < length + i;j++)

                {

                    remain = remain + sum[j % length];

                    if(remain < 0) 

                    {

                        flag = false;

                        i= j; //o(n^2)到o(n)的小小优化。若是从i到j个出现了负，那么前i ~ j不再可能作为起点

                        break;

                    }

                    else continue;

                }

                if(flag)

                    return i;

            }

        }

        return -1;

    }

};