[Java] LeetCode 485 : Max Consecutive Ones

LeetCode # 485

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Given a binary array, find the maximum number of consecutive 1s in this array.

 

 

Example 1:

 

Input: [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.

 

Note:

• The input array will only contain 0 and 1.
• The length of input array is a positive integer and will not exceed 10,000

 

 

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Solution #1 (1ms)

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int count = 0;
        int max = 0;

        for(int n : nums){
            if(n==1) {
                count ++;
                if(max < count) max = count;
            }
            else count = 0;
        }

        return max;
    }
}

 

Result #1

 

💡 간단하게 if 조건문을 이용하여 풀었다. 

 

 

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Solution #2

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int count = 0, max = 0;

        for (int n : nums) {
            max = Math.max(max,
                    count = n == 0 ? 0 : count + 1);
        }

        return max;
    }
}

 

Result #2

 

💡 Math 함수와 삼항 연산자를 사용하니 코드는 간결해졌다. 하지만 Math.max() 의 시간 복잡도는 O(N).  

따라서 시간이 2ms로 증가한 것을 확인할 수 있다.

 

 

 

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More Algorithm!

 

👇👇

 

github.com/ggujangi/ggu.leet-code

ggujangi/ggu.leet-code

 

 

 

 

출처 : leetCode