README_EN.md

485. Max Consecutive Ones

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Description

Given a binary array nums, return the maximum number of consecutive 1's in the array.

 

Example 1:

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.

Example 2:

Input: nums = [1,0,1,1,0,1]
Output: 2

 

Constraints:

• 1 <= nums.length <= 105
• nums[i] is either 0 or 1.

Solutions

Python3

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

Java

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

JavaScript

/**
 * @param {number[]} nums
 * @return {number}
 */
var findMaxConsecutiveOnes = function (nums) {
    let res = 0,
        t = 0;
    for (let num of nums) {
        if (num == 1) {
            ++t;
        } else {
            res = Math.max(res, t);
            t = 0;
        }
    }
    return Math.max(res, t);
};

TypeScript

function findMaxConsecutiveOnes(nums: number[]): number {
    let res = 0;
    let count = 0;
    for (const num of nums) {
        if (num === 0) {
            res = Math.max(res, count);
            count = 0;
        } else {
            count++;
        }
    }
    return Math.max(res, count);
}

Rust

impl Solution {
    pub fn find_max_consecutive_ones(nums: Vec<i32>) -> i32 {
        let mut res = 0;
        let mut count = 0;
        for num in nums {
            if num == 0 {
                res = res.max(count);
                count = 0;
            } else {
                count += 1;
            }
        }
        res.max(count)
    }
}

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