LeetCode重点题系列之【11. Container With Most Water】

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本文介绍了一种寻找能容纳最多水的两个垂直线的高效算法。通过使用双指针技术，从两端开始向中间搜索，根据高度较小的线进行移动，直至找到最优解。适用于至少包含两条线的数组。

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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.

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.

Example:

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

The logic is that we should use two pointers, one is on the very left, another is on the very right. If left heigh is smaller than right height, add left, else minus right. Maintain a maximal variable as the final result. The trick is the area equals to the smaller height of left and right multiply by the distance between them.

class Solution(object):
    def maxArea(self, height):
        """
        :type height: List[int]
        :rtype: int
        """
        l=0
        r=len(height)-1
        maximal=0
        while l<r:
            maximal=max(maximal,min(height[l],height[r])*(r-l))
            if height[l]<height[r]:
                l+=1
            else:
                r-=1
        return maximal