Leetcode11. The container with the most water-double pointer

11. The container with the most water

Difficulty: medium

Give you n non-negative integers a ₁ , a _2, …, an _{, and} each number represents a point ( i , a _i ) in the coordinates . Draw n vertical lines in the coordinates. The two end points of the vertical line i are ( i , a _i ) and ( i , 0). Find two of the lines so that the container that they form with the x- axis can hold the most water.

Explanation : You cannot tilt the container, and the value of n is at least 2.

The vertical line in the figure represents the input array [1,8,6,2,5,4,8,3,7]. In this case, the maximum value that the container can hold water (shown as the blue part) is 49.

Example:

输入：[1,8,6,2,5,4,8,3,7]
输出：49

Solution

Double pointers are used here, why can they be used?

1. The starting position of the double pointer specifies the search range. The left pointer starts at 0 and the right pointer is len(height)-1, which includes all possible positions .
2. Each time you move the pointer, move the corresponding one with the smaller height, and record the water holding volume before each movement. Note : Assuming that the left and right pointers are x and y respectively, height[x]<height[y], so you should move the left pointer next time. This means that height[x] can no longer be used as a new container wall, because if x does not move and the right pointer moves to the left, the new water volume must be smaller than the original . So this is correct.

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