485. Max Consecutive Ones

LeetCode easy original: C# #array #csharp #easy #leetcode #math #search

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

LeetCode Static

Дан бинарный 배열 nums, return максимальное количество последовательных единиц в 배열е.

예제:

Input: nums = [1,1,0,1,1,1]

Output: 3

Explanation: The first two digits or the last three digits are consecutive 1s. The maximum number of consecutive 1s is 3.

C# 해법

매칭됨/원본

public class Solution {
    public int FindMaxConsecutiveOnes(int[] nums) {
        int count = 0;
        int maxCount = 0;
        foreach (int num in nums) {
            if (num == 1) {
                count += 1;
            } else {
                maxCount = Math.Max(maxCount, count);
                count = 0;
            }
        }
        return Math.Max(maxCount, count);
    }
}

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 FindMaxConsecutiveOnes(vector<int>& nums) {
        int count = 0;
        int maxCount = 0;
        foreach (int num in nums) {
            if (num == 1) {
                count += 1;
            } else {
                maxCount = max(maxCount, count);
                count = 0;
            }
        }
        return max(maxCount, count);
    }
}

Java 해법

매칭됨/원본

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int count = 0;
        int maxCount = 0;
        for (int num : nums) {
            if (num == 1) {
                count += 1;
            } else {
                maxCount = Math.max(maxCount, count);
                count = 0;
            }
        }
        return Math.max(maxCount, count);
    }
}

JavaScript 해법

매칭됨/원본

var findMaxConsecutiveOnes = function(nums) {
    let count = 0
    let maxCount = 0
    for (let num of nums) {
        if (num === 1) {
            count += 1
        } else {
            maxCount = Math.max(maxCount, count)
            count = 0
        }
    }
    return Math.max(maxCount, count)
}

Python 해법

매칭됨/원본

class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        count = 0
        maxCount = 0
        for num in nums:
            if num == 1:
                count += 1
            else:
                maxCount = max(maxCount, count)
                count = 0
        return max(maxCount, count)

Go 해법

매칭됨/원본

func findMaxConsecutiveOnes(nums []int) int {
    count := 0
    maxCount := 0
    for _, num := range nums {
        if num == 1 {
            count++
        } else {
            if count > maxCount {
                maxCount = count
            }
            count = 0
        }
    }
    if count > maxCount {
        maxCount = count
    }
    return maxCount
}

Algorithm

Поддерживайте счетчик для подсчета единиц и увеличивайте его на 1 при встрече единицы.

Когда встречаете ноль, используйте текущий счетчик единиц для нахождения максимального количества последовательных единиц на данный момент, затем сбросьте счетчик единиц на 0.

В конце return максимальное значение.

😎

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