LeetCode: 134. Gas Station

LeetCode: 134. Gas Station
タイトルの説明
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.

問題を解く構想——シミュレーション
要求に応じて、各stationを起点として、プロセス全体を順次シミュレートする.
最適化:例えば、A,B,C,D,E,F,Gいくつかの駅は、B駅を起点とし、travelがF駅に着いたとき、ガソリンが足りなかった.では、B,C,D,E,Fのいくつかの駅を起点として、B駅に着くことはできません.なぜなら、C駅がDに着くことができる以上、Eに着くと、0に等しいガソリン量になるに違いないからだ.このとき、C,D,Eから出発すると、ガソリン量は実際にはこれらの駅から直接出発するガソリン量以上である.したがって、B駅からF駅に着けない以上、C,D,E駅にも行けない.したがって,F,C,D,Eのいくつかの局を直接スキップし,F局を起点としてシミュレーション計算を行うことができる.
ACコード
未最適化コード(Runtime:298 ms,complexity:O(n^2))

class Solution
{
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) 
    {
        int startStation = -1; //        
        int curStation;        //    travel   station
        int gasTank;           // gas tank     gas

        //       station     
        for(size_t i = 0; i < gas.size(); ++i)
        {
            //    : gas tank    gas   0，    station      i
            gasTank = 0;
            curStation = i;

            do
            {
                //   
                gasTank += gas[curStation];    

                //      
                gasTank -= cost[curStation];  
                curStation = (curStation+1) % gas.size();

                //        gas   ,       station     
                if(gasTank < 0)
                {
                    break;
                }
            }while(curStation != i);

            //    curStation     ，    gas tank     gas    ，         
            if(curStation == i && gasTank >= 0)
            {
                startStation = i;
                break;
            }
        }

        return startStation;
    }
};

最適化コード(Runtime:6 ms,complexity:O(n))

class Solution
{
public:
    int canCompleteCircuit(vector<int>& gas, vector<int>& cost) 
    {
        int startStation = -1; //        
        int travelStation;     //     travel   station  
        int curStation;        //   travel   station
        int gasTank;           // gas tank     gas

        //       station     (             station)
        for(size_t i = 0; i < gas.size(); i += travelStation)
        {
            //    : gas tank    gas   0，    station      i
            gasTank = 0;
            travelStation = 0;
            curStation = i;

            do
            {   
                //   
                gasTank += gas[(curStation)];    

                //      
                gasTank -= cost[curStation];  
                curStation = (curStation+1) % gas.size();
                ++travelStation;

                //        gas   ,       station     
                if(gasTank < 0)
                {
                    break;
                }
            }while(curStation != i);

            //    curStation     ，    gas tank     gas    ，         
            if(curStation == i && gasTank >= 0)
            {
                startStation = i;
                break;
            }
        }

        return startStation;
    }
};