leetcode11 Container With Most Water

Posted 2021-03-08 yawenw

 1 """
 2 Given n non-negative integers a1, a2, ..., an , where each represents a point at coordinate (i, ai). n vertical lines are drawn such that the two endpoints of line i is at (i, ai) and (i, 0). Find two lines, which together with x-axis forms a container, such that the container contains the most water.
 3 Note: You may not slant the container and n is at least 2.
 4 The above vertical lines are represented by array [1,8,6,2,5,4,8,3,7]. In this case, the max area of water (blue section) the container can contain is 49.
 5 Example:
 6 Input: [1,8,6,2,5,4,8,3,7]
 7 Output: 49
 8 """
 9 """
10 解法一：暴力解法
11 对数据量大，超时
12 Time Limit Exceeded
13 """
14 class Solution1:
15     def maxArea(self, height):
16         most_water = 0
17         if len(height) < 2:
18             return 0
19         for i in range(len(height)-1):
20             for j in range(i+1, len(height)):
21                 h = min(height[i], height[j])
22                 most_water = max(most_water, h*(j - i))
23         return most_water
24 
25 """
26 解法二：双指针法
27 让i=0, j=len(height)-1,从而保证底最长
28 如果高nums[i] < nums[j]  i++
29 否则j--
30 当重合i == j 退出迭代
31 """
32 class Solution2:
33     def maxArea(self, height) -> int:
34         most_water = 0
35         if len(height) < 2:
36             return 0
37         i = 0
38         j = len(height) - 1
39         while i < j:
40             h = min(height[i], height[j])
41             most_water = max(most_water, h*(j-i))
42             if height[i] < height[j]:
43                 i += 1
44             else:
45                 j -= 1
46         return most_water