Array force buckle (continuously updated)

Sum of two numbers （7.21）

Execution time ：3148ms
Write your own code ：

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        n = len(nums)
        for i in range(n):
            for j in range(i+1,n):
                if nums[i] + nums[j] == target:
                    return [i,j]

Official code ：

class Solution:
    def twoSum(self, nums: List[int], target: int) -> List[int]:
        hashtable = dict()
        for i, num in enumerate(nums):
            if target - num in hashtable:
                return [hashtable[target - num], i]
            hashtable[nums[i]] = i
        return []

Container for the most water （7.22 Medium difficulty ）

The area depends on the short board .① Therefore, even if a longer plate is encountered when the long plate moves inward , The area of the rectangle will not change ; When you encounter a shorter board , The area will become smaller .② So you want the area to get bigger , Only the short board can move inward ( Because the moving direction is fixed ), Of course, it is also possible to make the area smaller , But only in this way can there be the possibility of making the area larger

class Solution:
    def maxArea(self, height: List[int]) -> int:
        l, r = 0, len(height) - 1
        ans = 0
        
        while l < r:
            area = min(height[l], height[r]) * (r - l)
            ans = max(ans, area)
            if height[l] <= height[r]:
                l += 1
            else:
                r -= 1
      
        return ans

Remove duplicate items from an ordered array （7.22 Simple ）

reflection ：
Or use double pointers

class Solution:
    def removeDuplicates(self, nums: List[int]) -> int:
        n = len(nums)
        fast = slow = 1
        while fast < n:
            if nums[fast] != nums[fast - 1]:
                nums[slow] = nums[fast]
                slow += 1
            fast += 1
        
        return slow