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0485.Max Consecutive Ones

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简单
数组

English Version

题目描述

给定一个二进制数组 nums , 计算其中最大连续 1 的个数。

 

示例 1:

输入:nums = [1,1,0,1,1,1]
输出:3
解释:开头的两位和最后的三位都是连续 1 ,所以最大连续 1 的个数是 3.

示例 2:

输入:nums = [1,0,1,1,0,1]
输出:2

 

提示:

  • 1 <= nums.length <= 105
  • nums[i] 不是 0 就是 1.

解法

方法一:一次遍历

遍历数组,记录当前连续 1 的个数 cnt,以及最大连续 1 的个数 ans。如果当前元素为 1 ,则 cnt++,否则更新 ans,并且 cnt=0。最后返回 max(ans, cnt) 即可。

时间复杂度 O ( n ) ,空间复杂度 O ( 1 ) 。其中 n 为数组 nums 的长度。

Python3

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

Java

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

C++

class Solution {
public:
    int findMaxConsecutiveOnes(vector<int>& nums) {
        int cnt = 0, ans = 0;
        for (int v : nums) {
            if (v == 1) {
                ++cnt;
            } else {
                ans = max(ans, cnt);
                cnt = 0;
            }
        }
        return max(ans, cnt);
    }
};

Go

func findMaxConsecutiveOnes(nums []int) int {
	ans, cnt := 0, 0
	for _, v := range nums {
		if v == 1 {
			cnt++
		} else {
			ans = max(ans, cnt)
			cnt = 0
		}
	}
	return max(ans, cnt)
}

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)
    }
}

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);
};

PHP

class Solution {
    /**
     * @param Integer[] $nums
     * @return Integer
     */
    function findMaxConsecutiveOnes($nums) {
        $tmp = $max = 0;
        for ($i = 0; $i < count($nums); $i++) {
            if ($nums[$i] == 1) {
                $tmp++;
            } else {
                $max = max($tmp, $max);
                $tmp = 0;
            }
        }
        return max($tmp, $max);
    }
}