Given n non-negative integers a1a2, ..., an, where each represents a point at coordinate (iai). n vertical lines are drawn such that the two endpoints of line i is at (iai) 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.

【思路1】

暴力拆解,找出所有的组合,返回其中最大的。但是这样运行会超时,代码如下:

 public class Solution {
public int maxArea(int[] height) {
if(height == null || height.length == 0) return 0;
int max = 0;
for(int i = 0; i < height.length - 1; i++){
for(int j = i + 1; j < height.length; j++){
int minH = Math.min(height[i], height[j]);
max = Math.max(max, (j - i)*minH);
}
}
return max;
}
}

【思路2】

The brute force solution can definitely lead us to the right answer just by doing too many redundant comparisons. When two pointer approach comes to mind, it is intuitive to set both pointers i, j at each end of this array, and move them strategically to the middle of array, update the answer during this process return the answer when we reach the end of array. Suppose now we have the scenarios below:

7, 5, 6, 9

i        j

When i = 1, j = 4,

ans = min(7, 9) * (4 - 1) = 21

What's next? Should we move i or j? We notice that to calculate the area, the height is really identified by the smaller number / shorter end between the two ends, since it's required that you may not slant the water, so it sounds like Bucket theory: how much water a bucket can contain depends on the shortest plank. So, as to find the next potential maximum area, we disregard the shorter end by moving it to the next position. So in the above case, the next status is to move i to the left,

7, 5, 6, 9

   i     j

update:

area (i, j) = area(2, 4) = min(5, 9) * (4 - 2) = 10
ans = max(21, 10) = 21

You may notice that, if we move j instead, you actually get a larger area for length of 2:

area (i, j) = area(1, 3) = min(7, 6) * (3 - 1) = 18

Does that mean this approach will not work? If you look at this way, we move pointer as to get the next potential max, so it doesn't need to be the maximum for all combinations with length l. Even though 18 is greater than 10, it's smaller than 21 right? So don't worry, we can move on to find the next potential maximum result. Now we need to prove, why disregard the shorter end can safely lead us to the right answer by doing a little maths.

Given an array: a1, a2, a3, a4, ai, ......, aj, ......, an
i j

Assume the maximum area so far is ans, we prove that

"By moving shorter end pointer further doesn't eliminate the final answer (with two ends at maxi, maxj respectively) in our process"

Suppose we have two ends at (i, j) respectively at this moment:

(i) If the final answer equals what we have already achieved, it's done! In this scenario, we must have

maxi <= i, maxj >= j

(ii) Otherwise, we know as we move any pointer further, the length of the next rectangle decreases, so the height needs to increase as to result in a larger area. So we have

min(height[maxi], height[maxj]) > min(height[i], height[j])

So the smaller one in height[i], height[j] won't become any end in the maximum rectangle, so it's safe to move forward without it.

Till now, it has been proved that this approach can work in O(n) time since we advance one end towards the middle in each iteration, and update ans takes constant time in each iteration.

代码如下:

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

LeetCode OJ 11. Container With Most Water的更多相关文章

  1. 《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 ...

  2. 【LeetCode】11. Container With Most Water 盛最多水的容器

    作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 个人公众号:负雪明烛 本文关键词:盛水,容器,题解,leetcode, 力扣,python ...

  3. 【LeetCode】11. Container With Most Water

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

  4. leetcode problem 11 Container With Most Water

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

  5. Leetcode Array 11 Container With Most Water

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

  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 Array Medium 11. Container With Most Water

    Description Given n non-negative integers a1, a2, ..., an , where each represents a point at coordin ...

  9. leetcode面试准备:Container With Most Water

    leetcode面试准备:Container With Most Water 1 题目 Given n non-negative integers a1, a2, ..., an, where eac ...

随机推荐

  1. vedio_note_1

    同步复位 always @ (posedge clk) ....... 异步复位 always @ (posedge clk or negedge rst_n) ....... 异步复位和同步复位的优 ...

  2. android 控件注意点

    控件一:listview 问题一:当listview的item中存在按钮这种控件时 item点击不能响应问题? 解决方案:在item的自定义控件的最外层空间 上添加属性 android:descend ...

  3. moodle其他代码

    , $sectionnum=false, $strictness=IGNORE_MISSING):给课程模块一个id,找出coursemoudle的描述 get_coursemodule_from_i ...

  4. VS2012及以上版本 程序打包部署详解

    引用:  http://blog.csdn.net/zhang_xinxiu/article/details/9099757 程序编写测试完成后接下来我们要做的是打包部署程序,但VS2012让人心痛的 ...

  5. c#弱事件(weak event)

    传统事件publisher和listener是直接相连的,这样会对垃圾回收带来一些问题,例如listener已经不引用任何对象但它仍然被publisher引用 垃圾回收器就不能回收listener所占 ...

  6. 每天点滴的进行,css+div简单布局...布局

    <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <title> ...

  7. 按键精灵*ff

    Function gethttp(URL) Set objXML=CreateObject("Microsoft.XMLHTTP") objXML.Open "Get&q ...

  8. MSSQL2008 中文乱码问题 (引自ljg888的专栏)

    PHP向MSSQL2008中写入数据,中文乱码   首先:查看SQLserver编码格式的SQL语句为:     SELECT  COLLATIONPROPERTY('Chinese_PRC_Stro ...

  9. 窗口、easyui-window、easyui-panel、easyui-linkbutton

    //窗口 <script type="text/javascript" src="js/jquery.min.js"></script> ...

  10. Mac Java maven环境变量

    p.p1 { margin: 0.0px 0.0px 0.0px 0.0px; font: 11.0px Menlo; color: #000000; background-color: #fffff ...