134. Gas Station

LeetCode medium original: C# #array #csharp #leetcode #medium

선택한 UI 언어에 맞게 문제 텍스트를 러시아어에서 번역합니다. 코드는 변경하지 않습니다.

LeetCode Static

Вдоль кругового маршрута расположены n заправочных станций, на каждой из которых находится определённое количество топлива gas[i].

У вас есть автомобиль с неограниченным топливным баком, и для проезда от i-й станции к следующей (i + 1)-й станции it is required cost[i] топлива. Путешествие начинается с пустым баком на одной из заправочных станций.

C# 해법

매칭됨/원본

public class Solution {
    public int CanCompleteCircuit(int[] gas, int[] cost) {
        int currGain = 0, totalGain = 0, answer = 0;
        for (int i = 0; i < gas.Length; ++i) {
            totalGain += gas[i] - cost[i];
            currGain += gas[i] - cost[i];
            if (currGain < 0) {
                answer = i + 1;
                currGain = 0;
            }
        }
        return totalGain >= 0 ? answer : -1;
    }
}

C++ 해법

자동 초안, 제출 전 검토

#include <bits/stdc++.h>
using namespace std;

// Auto-generated C++ draft from the C# solution. Review containers, LINQ and helper types before submit.
class Solution {
public:
    public int CanCompleteCircuit(vector<int>& gas, vector<int>& cost) {
        int currGain = 0, totalGain = 0, answer = 0;
        for (int i = 0; i < gas.size(); ++i) {
            totalGain += gas[i] - cost[i];
            currGain += gas[i] - cost[i];
            if (currGain < 0) {
                answer = i + 1;
                currGain = 0;
            }
        }
        return totalGain >= 0 ? answer : -1;
    }
}

Java 해법

매칭됨/원본

class Solution {
    public int canCompleteCircuit(int[] gas, int[] cost) {
        int currGain = 0, totalGain = 0, answer = 0;

        for (int i = 0; i < gas.length; ++i) {
            totalGain += gas[i] - cost[i];
            currGain += gas[i] - cost[i];
            if (currGain < 0) {
                answer = i + 1;
                currGain = 0;
            }
        }

        return totalGain >= 0 ? answer : -1;
    }
}

JavaScript 해법

매칭됨/원본

var canCompleteCircuit = function (gas, cost) {
    let currGain = 0,
        totalGain = 0,
        answer = 0;
    for (let i = 0; i < gas.length; ++i) {
        totalGain += gas[i] - cost[i];
        currGain += gas[i] - cost[i];
        if (currGain < 0) {
            answer = i + 1;
            currGain = 0;
        }
    }
    return totalGain >= 0 ? answer : -1;
};

Python 해법

매칭됨/원본

class Solution:
    def canCompleteCircuit(self, gas: List[int], cost: List[int]) -> int:
        total_gain = 0
        curr_gain = 0
        answer = 0

        for i in range(len(gas)):
            total_gain += gas[i] - cost[i]
            curr_gain += gas[i] - cost[i]
            if curr_gain < 0:
                curr_gain = 0
                answer = i + 1

        return answer if total_gain >= 0 else -1

Go 해법

매칭됨/원본

func canCompleteCircuit(gas []int, cost []int) int {
    var currGain int = 0
    var totalGain int = 0
    var answer int = 0
    for i := 0; i < len(gas); i++ {
        totalGain += gas[i] - cost[i]
        currGain += gas[i] - cost[i]
        if currGain < 0 {
            answer = i + 1
            currGain = 0
        }
    }
    if totalGain >= 0 {
        return answer
    } else {
        return -1
    }
}

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