基础题数组-485、283、27-526互联

485. 最大连续 1 的个数

 1 class Solution:
 2     def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
 3         maxCount = count = 0
 4 
 5         for i, num in enumerate(nums):
 6             if num == 1:
 7                 count += 1
 8             else:
 9                 maxCount = max(maxCount, count)
10                 count = 0
11         
12         maxCount = max(maxCount, count)
13         return maxCount

        count = 0 
        ans = 0
        for i in nums:
            if i == 1:
                count += 1
                if ans < count:
                    ans = count
            else:
                count = 0
        return ans

       #先化为str,再用split方法将其以"0"分隔，结果便是list中最长的item的长度
        s = ""

        for i in nums:
            s += str(i)

        L2 = s.split("0")
        return max([len(i) for i in L2])

        #快慢同向双指针（移动窗口法）
        slow = fast = 0
        n = len(nums)
        max_count = 0
        while slow < n:
            # 找出连续1中，第一个1的位置
            if nums[slow] != 1:
                slow += 1
            else:
                # fast指针从slow指针开始遍历，直至nums[fast]!=1
                fast = slow
                while fast < n and nums[fast] == 1:
                    fast += 1
                # 局部最长fast-slow和全局最长max_count取最大值
                max_count = max(max_count, fast - slow)
                # 更新slow指针的位置到fast指针处，继续循环，寻找下一个1
                slow = fast
        return max_count

tournamsan tennis 283e cow