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

自己的解法

判断连续的1，设置一个计数器，如果数组中的元素为1的计数器自增，然后将计数器的值赋值给连续1末尾的数组值。如果数组中的元素等于0的话，计数器清0，继续循环判断。最后判断数组中的最大值就是最大连续1的个数。
缺点：时间复杂度过长，每次要遍历2遍数组。

 public int findMaxConsecutiveOnes(int[] nums) {
         
        int length = nums.length;
        int count = 0;
        int max = nums[0];
        for (int i = 0; i <= length - 1; i++) {

            if (nums[i] == 1) {
                count++;
                nums[i] = count;
            } else if (nums[i] == 0) {
                count = 0;
                continue;
            }
        }

        for (int j = 0; j <= length - 1; j++) { 
            if (nums[j] > max) {
                max = nums[j];
            }
        }
        return max;      
}

优化解法

Java中提供判断最大值的函数，Math.max(a,b)，直接使用；不用自己遍历数组寻找最大值。

  public int findMaxConsecutiveOnes(int[] nums) {
        int result = 0;
        int count = 0;
        
        for (int i = 0; i < nums.length; i++) {
            if (nums[i] == 1) {
            count++;
            result = Math.max(count, result);
            }
            else count = 0;
        }
        
        return result;
    }
}