leetCode-Max Consecutive Ones

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

Example1:

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

My Solution:

class Solution {
    public int findMaxConsecutiveOnes(int[] nums) {
        int count = 0;
        int maxN = 0;
        for(int i : nums){
            if(i == 1){
                count++;
                 if(count > maxN){
                    maxN = count;
                }
            }else{

                count = 0;
            }
        }
        return maxN;
    }
}

改进:

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

        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 1) {
                count++;
                //替换为了Math.max()函数，更简洁
                result = Math.max(count, result);
            }
            else count = 0;
        }

        return result;
    }
}

总结:
注意每次碰到1的时候增加count,然后求max(count,maxN)，然后每次num = 0将count置0就可以了。