原题是 ：

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

思路是 ：

用一个变量num来记录1的个数，遇到0时num归0；用另一个变量ans来记录最大的nums，注意最后一步也要再取一次最大，否则只是遇到0才取最大，如果最长1序列出现在末尾，会疏忽最后一步对ans最大取值的更新。

代码是:

class Solution(object):
    def findMaxConsecutiveOnes(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        num,ans = 0,0
        for i in xrange(len(nums)):
            if nums[i] == 1:
                num = num + 1
            elif nums[i] == 0:
                ans = max(num,ans)
                num = 0
        ans = max(num,ans)
        return ans

借鉴别人的代码

class Solution(object):
    def findMaxConsecutiveOnes(self, nums):
        cnt = 0
        ans = 0
        for num in nums:
            if num == 1:
                cnt += 1
                ans = max(ans, cnt)
            else:
                cnt = 0
        return ans

一个是循环变量的设置比我的简单，另外在遇到1的时候就更新最大值，避免了最后出现问题的可能性。