public class Solution {
public int maxArea(int[] height) {
int left = 0, right = height.length - 1;
int maxArea = 0; while (left < right) {
maxArea = Math.max(maxArea, Math.min(height[left], height[right]) * (right - left));
if (height[left] < height[right])
left++;
else
right--;
} return maxArea;
} }

[Leetcode]011. Container With Most Water的更多相关文章

  1. 【JAVA、C++】LeetCode 011 Container With Most Water

    Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, ai). ...

  2. 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 <= ...

  3. No.011 Container With Most Water

    11. Container With Most Water Total Accepted: 86363 Total Submissions: 244589 Difficulty: Medium Giv ...

  4. LeetCode--No.011 Container With Most Water

    11. Container With Most Water Total Accepted: 86363 Total Submissions: 244589 Difficulty: Medium Giv ...

  5. LeetCode OJ Container With Most Water 容器的最大装水量

    题意:在坐标轴的x轴上的0,1,2,3,4....n处有n+1块木板,长度不一,任两块加上x轴即可构成一个容器,其装水面积为两板的间距与较短板长之积,以vector容器给出一系列值,分别代表在0,1, ...

  6. 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 ...

  7. 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 ...

  8. 【LeetCode】011 Container With Most Water

    题目: Given n non-negative integers a1, a2, ..., an, where each represents a point at coordinate (i, a ...

  9. [LeetCode][Python]Container With Most Water

    # -*- coding: utf8 -*-'''https://oj.leetcode.com/problems/container-with-most-water/ Given n non-neg ...

随机推荐

  1. 用nginx搭建http/rtmp/hls协议的MP4/FLV流媒体服务器

    前前后后搭建了两三个星期,终于可以告一段落,nginx实在是有点强大.写一篇笔记来记录一下这个过程中的思路和解决方案. 一.搭建nginx平台: 基本是基于http://blog.csdn.net/x ...

  2. NYOJ-求逆序数 ----------------待解决,WA

    描述 在一个排列中,如果一对数的前后位置与大小顺序相反,即前面的数大于后面的数,那么它们就称为一个逆序.一个排列中逆序的总数就称为这个排列的逆序数. 现在,给你一个N个元素的序列,请你判断出它的逆序数 ...

  3. unity渲染层级关系小结

    http://blog.csdn.net/meegomeego/article/details/42060389 最近连续遇到了几个绘制图像之间相互遮挡关系不正确的问题,网上查找的信息比较凌乱,所以这 ...

  4. bzoj 1101 Zap —— 莫比乌斯反演

    题目:https://www.lydsy.com/JudgeOnline/problem.php?id=1101 直接莫比乌斯反演. 代码如下: #include<cstdio> #inc ...

  5. Poj 1887 Testing the CATCHER(LIS)

    一.Description A military contractor for the Department of Defense has just completed a series of pre ...

  6. Poj 1458 Common Subsequence(LCS)

    一.Description A subsequence of a given sequence is the given sequence with some elements (possible n ...

  7. HUD1455:Sticks

    Sticks Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Subm ...

  8. POJ3630(Trie树)

    Phone List Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 26385   Accepted: 7957 Descr ...

  9. 第三课 go语言基础语法

    http://www.runoob.com/go/go-basic-syntax.html 1 行分隔符 在 Go 程序中,一行代表一个语句结束.每个语句不需要像 C 家族中的其它语言一样以分号 ; ...

  10. 【转】ruby 时间日期处理

    我们可以使用Time类来生成一个当前时间的对象: t = Time.new 或 t = Time.now Time类有类方法mktime(同义方法是local方法)来根据传入的参数生成时间对象,并且它 ...