[LC] 11. Container With Most Water
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
class Solution {
public int maxArea(int[] height) {
int left = 0, right = height.length - 1;
int res = 0;
while (left < right) {
res = Math.max(res, Math.min(height[left], height[right]) * (right - left));
// move the smaller slant
if (height[left] < height[right]) {
left += 1;
} else {
right -= 1;
}
}
return res;
}
}
[LC] 11. Container With Most Water的更多相关文章
- leetcode 11. Container With Most Water 、42. Trapping Rain Water 、238. Product of Array Except Self 、407. Trapping Rain Water II
11. Container With Most Water https://www.cnblogs.com/grandyang/p/4455109.html 用双指针向中间滑动,较小的高度就作为当前情 ...
- Leetcode 11. Container With Most Water(逼近法)
11. Container With Most Water Medium Given n non-negative integers a1, a2, ..., an , where each repr ...
- LeetCode Array Medium 11. Container With Most Water
Description Given n non-negative integers a1, a2, ..., an , where each represents a point at coordin ...
- 刷题11. Container With Most Water
一.题目说明 11.Container With Most Water,这个题目难度是Medium. 二.我的做法 乍一看,简单啊,两个for循环就可以了,我在本地写的. #include<io ...
- 如何装最多的水? — leetcode 11. Container With Most Water
炎炎夏日,还是呆在空调房里切切题吧. Container With Most Water,题意其实有点噱头,简化下就是,给一个数组,恩,就叫 height 吧,从中任选两项 i 和 j(i <= ...
- [leecode]---11.container with most water
description: Input: [1,8,6,2,5,4,8,3,7]Output: 49 思路1: 从(1,a1)开始向后算面积,需要两层n循环,时间复杂度n2 思路2: 找出数组中最大的数 ...
- 《LeetBook》leetcode题解(11):Container With Most Water[M] ——用两个指针在数组内移动
我现在在做一个叫<leetbook>的免费开源书项目,力求提供最易懂的中文思路,目前把解题思路都同步更新到gitbook上了,需要的同学可以去看看 书的地址:https://hk029.g ...
- LeetCode 11. Container With Most Water (装最多水的容器)
Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). ...
- 11. Container With Most Water 装水最多的容器
[抄题]: Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ...
随机推荐
- HALCON形状匹配讲解
HALCON形状匹配讲解 https://blog.csdn.net/linnyn/article/details/50663328 https://blog.csdn.net/u014608071/ ...
- Thread--生产者消费者假死分析
package p_c_allWait; public class ValueObject { public static String value = ""; } package ...
- ModernCNN
# 深度卷积神经网络(AlexNet) LeNet: 在大的真实数据集上的表现并不尽如⼈意. 1.神经网络计算复杂. 2.还没有⼤量深⼊研究参数初始化和⾮凸优化算法等诸多领域. 机器学习的特征提取:手 ...
- 装饰器参数加log
#有多个函数,需要计算他们的执行时间,加logimport timedef logger(flag): def show_time(f): #foo对象 def inner(*x,**y): star ...
- Spring事务管理 —— readOnly只读事务
事务是什么?事务是一个原子操作,由一系列动作组成.事务的原子性确保动作要么全部完成,要么完全不起作用. 下面来看一个项目中遇到的问题: 有这么一个需求,我们要查询一些数据,但是在查询这个数据之前我们要 ...
- mysql 不停机 短时间锁表 备份 主备同步 新增备份机器
刷新数据 [root@localhost ~]# mysql -e 'flush tables with read lock;' 锁表刷新表数据 [root@localhost ~]# mys ...
- iPhoneX的后遗症要持续多久?
iPhone X的推出算得上苹果历史上的大事件,这款梳着刘海头型的手机作为iPhone十周年纪念款手机,承载着苹果和整个产业链巨大的希望,正因如此,包括苹果在内的大量企业,把宝都压到了这款手机上.后来 ...
- Bless All
# php code $i = 2333 $myJXOI = JXOI() while($i == 2333){ ++myJXOI.score , ++myJXOI.rp , --myJXOI.常数 ...
- 关于SG函数
Sprague-Grundy定理(SG定理): 游戏和的SG函数等于各个游戏SG函数的Nim和.这样就可以将每一个子游戏分而治之,从而简化了问题.而Bouton定理就是Sprague-Grundy定理 ...
- IOC&AOP