# Leetcode # container with most water (Todo)

Source: Internet
Author: User

Description:

Implementation 1 -- evaluate all possible values, O (N ^ 2), timeout (because the timeout did not run all test cases, so there is no other problem)

Code:

` 1     def maxArea(self, height): 2         tmp = len(height) 3         if tmp == 0 or tmp == 1: return 0 4         if tmp == 2: return abs(height[1] - height[0]) 5  6         minus_lst = [height[i] - height[i-1] for i in range(1, len(height))] 7         # len(minus_lst) >= 2 8  9         dis, max, l = 2, 0, len(minus_lst)10 11         while True:12             for i in range(l):13                 if i + dis > l: break14 15                 area = abs(sum(minus_lst[i:i+dis]) * dis)16                 if area > max: max = area 17 18             dis += 119             if dis > l: break20 21         return max`

Implementation 2:

For each node, take the location of the node longer than him. At this time, Area = Distance Difference * Height of the node, practice complexity O (N ^ 2), timeout

Code:

` 1     def maxArea(self, height): 2         tmp = len(height) 3         result = 0 4  5         for i in range(tmp): 6             for j in range(tmp): 7                 if height[j] >= height[i]: 8                     area = height[i] * abs(i - j) 9                     if area > result: result = area10 11         return result`

Optimization-1 still times out

` 1     def maxArea(self, height): 2         tmp = len(height) 3         result = 0 4  5         record = [] 6         for i in range(tmp): 7             for j in range(tmp): 8                 if height[j] >= height[i]: 9                     record.append(abs(j - i))10             area = height[i] * max(record)11             if area > result: result = area12             record = []13 14         return result`

Optimization-2 timeout

` 1     def maxArea(self, height): 2         tmp = len(height) 3         result = 0 4  5         for i in range(tmp): 6             t = None 7             for j in range(i): 8                 if height[j] > height[i]: 9                     t = abs(j - i)10                     break11 12             if t is not None:13                 area_1 = t * height[i]14                 if area_1 > result: result = area_115 16             t = None17             for j in range(i, tmp)[::-1]:18                 if height[j] > height[i]:19                     t = abs(j - i)20                     break21 22             if t is not None:23                 area_2 = t * height[i]24                 if area_2 > result: result = area_225 26         return result`

Implementation 3

Starting from the left and right endpoints, greedy, constantly pushing small values, intuition, how to prove that greedy can be used

When J is on the left and I is on the right, there will be no larger relationship between IJ. What are the two sides?

Todo

Related Keywords:

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