[LeetCode] 134. Gas Station - 문제풀이

Description

There are n gas stations along a circular route, where the amount of gas at the ith station is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from the ith station to its next (i + 1)th station. You begin the journey with an empty tank at one of the gas stations.

Given two integer arrays gas and cost, return the starting gas station's index if you can travel around the circuit once in the clockwise direction, otherwise return -1. If there exists a solution, it is guaranteed to be unique

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.

Constraints:

• gas.length == n
• cost.length == n
• 1 <= n <= 10^5
• 0 <= gas[i], cost[i] <= 10^4

Solution 1. Greedy

public int canCompleteCircuit(int[] gas, int[] cost) {
    //1.
    int len = gas.length;
    int sum = 0;
    int result = 0;
    int total = 0;
    for (int i = 0; i < len; i++) {
        sum += gas[i] - cost[i];
        if (sum < 0) {
            total += sum;
            sum = 0;
            result = i + 1;
        }
    }
    total += sum;
    return total < 0 ? -1 : result;

}

합계가 음수일때마다 시작점을 재설정하고 차를 새로 출발 시키고 마지막 시작점을 저장합니다. 그 동안 남은 가스를 모두 합산합니다. 마지막으로 음수이면 끝낼 수 없으므로 -1을 반환합니다.

**Solution 2. 전체 탐색(**Brute Force)

public int canCompleteCircuit2(int[] gas, int[] cost) {
    //2. Brute Force
    int len = gas.length;
    int tank = 0;
    for (int i = 0; i < len; i++) {

        int distance = len;
        int j = i;
        while(distance > 0){ //length 만큼 시계방향으로 돌아야함
            tank += gas[j];     // 충전
            if(tank < cost[j]){ // 더 이상 갈수가 없어
                break;
            }
            tank -= cost[j];
            j = j==len-1? 0 : j+1; //다음 정거장 index
            distance--;
        }

        if(distance==0){ // 다 돌고 나옴
            return i;
        }else{
            tank = 0;
        }
    }
    return -1;
}

전체 탐색을 통해 모든 케이스를 확인해 봅니다. 각 시작점(i) 부터 시계방향으로 한바퀴(distance)를 돌 수 있는 케이스를 발견하면 첫번째 가능한 인덱스를 반환해줍니다.

Reference

Gas Station - LeetCode