LeetCode - Gas Station

문제 설명

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

입출력 예시

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.

제약사항

gas.length == n
cost.length == n
1 <= n <= 104
0 <= gas[i], cost[i] <= 104

Solution #1 Brute Force

class Solution {

    public int canCompleteCircuit(int[] gas, int[] cost) {

        int stationLength = gas.length;

        for (int i = 0; i < stationLength; i++) {
            int remainGas = gas[i] - cost[i];
            if (remainGas >= 0) {
                int j = (i + 1) % stationLength;
                while (j != i) {
                    remainGas = remainGas + gas[j] - cost[j];
                    if (remainGas < 0) {
                        break;
                    }
                    j = (j + 1) % stationLength;
                }
                if (i == j) {
                    return i;
                }
            }
        }
        return -1;
    }
}

Solution #2

[전략]
1.for문을 순회하면서,
2.인덱스 0에서부터 시작하는 연속적인 마이너스 축적값을 계산하고
3.연속적인 합이 마이너스가 되는 지점 이후 축적된 플러스 축적값을 계산하여
4.순회가 끝나면 플러스 축적값 - 마이너스축적값을 뺀 값이 0이상이면 promise

    public int canCompleteCircuit(int[] gas, int[] cost) {

        int stationLength = gas.length;
        int preMinusSum = 0;
        int plusSum = 0;
        int startPoint = 0;
        for (int i = 0; i < stationLength; i++) {
            int curDiff = gas[i] - cost[i];
            if (plusSum + curDiff >= 0) {
                plusSum += curDiff;
            } else {
                preMinusSum += plusSum + curDiff;

                plusSum = 0;
                startPoint = (i + 1) % stationLength;
            }
        }

        if (preMinusSum + plusSum >= 0) {
            return startPoint;
        }
        return -1;
    }