基础题数组-485、283、27

485. 最大连续 1 的个数

class Solution:
    def findMaxConsecutiveOnes(self, nums: List[int]) -> int:
        maxCount = count = 0

        for i, num in enumerate(nums):
            if num == 1:
                count += 1
            else:
                maxCount = max(maxCount, count)
                count = 0
        
        maxCount = max(maxCount, count)
        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

 

 

 283. 移动零

class Solution:
    def moveZeroes(self, nums: List[int]) -> None:
        """
        Do not return anything, modify nums in-place instead.
        """
        #不交换，之间覆盖
        index = 0
        i = 0
        n = len(nums)
        while i < n :
            if nums[i] != 0:
                nums[index] = nums[i]
                index += 1
            i += 1

        #python 在 for 里修改索引变量不会影响到下一个循环，使用 while 才行
        while index < n: 
            nums[index] = 0
            index += 1
        
        return nums

27. 移除元素

class Solution:
    def removeElement(self, nums: List[int], val: int) -> int:
        #pop删除元素
        n = len(nums)
        i = 0
        if n == 0:
            return 0

        while i < n:
            if nums[i] == val:
                nums.pop(i)
                n -= 1 #数组长度发生变化
            else:
                i += 1 #不等于val值才往下走一位

        return len(nums)

        #双指针 更快
        s = f = 0

        while f < len(nums):
            if nums[f] != val:
                nums[s] = nums[f]
                s += 1
            f += 1
        
        return s