Container with the most water (double pointer)
11. Container with the most water
Medium difficulty 1342 Collection sharing Switch to English Follow feedback
You are given n non-negative integers a 1, a 2, ..., an , each number represents a point ( i , ai ) in coordinates . Draw n vertical lines in the coordinates. The two endpoints of the vertical line i are ( i , ai ) and ( i , 0). Find two of the lines so that the container they form with the x- axis can hold the most water.
Note: 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 (represented by the blue part) is 49.
Examples:
输入：[1,8,6,2,5,4,8,3,7]
输出：49
answer:
class Solution:
def maxArea(self, height: List[int]):
vol=0
leng=len(height)
l,r=height[0],height[leng-1]
nl,nr=0,leng-1
for i in range(leng-1):
mi=min(l, r)
vol = max(mi*(leng-i-1),vol)
if mi==l:
nl+=1
l=height[nl]
else:
nr-=1
r=height[nr]
return(vol)
Resolution practice:
After reading the question, the first thing I thought of was the double loop, but it felt very troublesome, so I set a different course. I may have done such a question or have been taught in an algorithm class before, and I always feel a little impressed.
I thought of moving on both sides, just selecting a small side to move, so the correct way is him ,,,, double pointer
Define two boundary variables r and l, and the cycle moves to the middle. In the process, compare the two volumes vol and select a large value and then assign it to vol. The boundary is updated after each cycle.
As for why you want to move the small side, please see here
