leetcode 485. Max Consecutive Ones

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

解法1: 贪心，看到1就一直计数。

class Solution(object):
    def findMaxConsecutiveOnes(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        ans = 0
        i = 0
        length = len(nums)
        while i<length:
            ones = 0
            while i<length and nums[i] == 1:
                i += 1
                ones += 1
            i += 1 # i == length or nums[i]==0
            ans = max(ones, ans)
        return ans

OR:

class Solution(object):
    def findMaxConsecutiveOnes(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        # greedy
        ans = 0
        i = 0
        length = len(nums)
        while i<length:
            while i<length and nums[i] == 0:
                i += 1
            if i==length: break
            ones = 0
            while i<length and nums[i] == 1:
                i += 1
                ones += 1
            ans = max(ones, ans)
        return ans

解法2:

使用计数器，看到1就+1，看到0就reset计数器为0.

class Solution(object):
    def findMaxConsecutiveOnes(self, nums):
        """
        :type nums: List[int]
        :rtype: int
        """
        ans = 0
        ones = 0
        for n in nums:
            if n == 0:
                ones = 0
            else:
                ones += 1
            ans = max(ones, ans)
        return ans