
正文
leetcode 【 Container With Most Water 】python 实现
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题目 :
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.
代码 :oj测试通过 Runtime: 132 ms
class Solution:
# @return an integer
def maxArea(self, height):
# none case or one element case
if height is None or len(height)<2:
return 0
# left point and right point
left = 0
right = len(height)-1
max_container = 0
while left<right :
curr_container = min(height[left],height[right]) * (right-left)
max_container = max(curr_container,max_container)
if height[left]>height[right]:
right = right-1
else:
left = left+1
return max_container
思路 :
数组前后双指针技巧。
有点儿像动态规划。
两个指针一左一右left right
面积为:min(height[left],height[right])*(right-left)
指针迭代条件为:哪边的指针所指位置的高度小,就从哪边往中间移动。每一步更新一次max_container的值。
为什么哪边的指针所指的位置高度小就从哪边往中间移动呢?能装多少水是有较短的那边决定的,因此如果寻求装更多的水,则应该优先从较短的一侧开始求变。
这样一来,每一次迭代后,都保证max_container保存了当前以及之前的可能最大蓄水量。







