【算法】leetcode 11 Conatainer With Most Water

題目：

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.

Note: You may not slant the container and n is at least 2.

舉例：
Input: [1,8,6,2,5,4,8,3,7]
Output: 49

 

題目分析：

       首先最先想到的是暴力法。遍歷所有組合，求出最大值。

class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        maxvalue = 0
        for i in range(len(height)):
            for j in range(i+1,len(height)):
                maxvalue = max( min(height[i], height[j]) * abs(i - j),maxvalue)
        return maxvalu

      此方法複雜度O(n^2)， 無法通過。

      進而利用兩端指針分析法，參考 https://blog.csdn.net/qq_36721548/article/details/80159570 , 即兩個指針分別置於首位，如果保留較小的長度，則後續遍歷的容器大小必然會小於當前值（寬度在減少）；因而保留較大值，移動較小值的指針。同理類推至重合。保留過程中的最大值。算法複雜度僅爲 O(n)。（runtime 36 ms）

class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        i,j,area = 0, len(height)-1, 0
        while i<=j: 
            if height[i] > height[j]:
                h = height[j]*(j-i)
                j -= 1
            else:
                h = height[i]*(j-i)
                i += 1
            area = max(area,h)
        return area