1004.Maximum continuity1Number III Maximum continuity1Number III

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        k_mean = k
        # Used to record how many arrays there are now
        flag = 0
        # Used to record the maximum land number
        max_flag = 0
        for start in range(len(nums)):
            tail = start
            while k >= 0 and tail <= len(nums) - 1:
                if nums[tail] == 1:
                    tail += 1
                    flag += 1
                elif nums[tail] == 0 and k > 0:
                    tail += 1
                    k -= 1
                    flag += 1
                elif nums[tail] == 0 and k == 0:
                    k = k_mean
                    max_flag = max(max_flag, flag)
                    flag = 0
                    break
                if tail == len(nums):
                    max_flag = max(max_flag, flag)
                    flag = 0
                    break
        return max_flag

class Solution:
    def longestOnes(self, nums: List[int], k: int) -> int:
        """
        Thinking：1. k=0It can be understood as the problem of the maximum duplication sub -string
             2. If the current window value-In the window1Number <= k: Then expand the window(right+1)
                If the current window value-In the window1Number > k: Swipe the window to the right(left+1)
        method：Hash table + Sliding window
        """
        n = len(nums)
        o_res = 0
        left = right = 0
        while right < n:
            if nums[right]== 1: o_res += 1
            if right-left+1- o_res > k:
                if nums[left]== 1: o_res -= 1
                left += 1
            right += 1
        return right - left