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
and1
. - The length of input array is a positive integer and will not exceed 10,000
思路:
把list通过map函数转为相应的字符串list,然后使用””.join()函数连成一个字符串,问题转为,通过”0″ split之后最长的子字符串的长度。
def findMaxConsecutiveOnes(self, nums):
"""
:type nums: List[int]
:rtype: int
"""
return len(max("".join(map(str, nums)).split("0")))
别人解法思路:
设置两个计数器:结果值和计数值。以此访问list,如果值为1,计数值加一,结果值取结果值和计数值中的大值。如果值为0,计数值复为0。
def findMaxConsecutiveOnes2(self, nums):
result = 0
count = 0
for i in range(len(nums)):
if nums[i] == 1:
count += 1
result = max(count, result)
else:
count = 0
return result