题目:

Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by array nums. You are asked to burst all the balloons. If the you burst balloon i you will get nums[left] * nums[i] * nums[right] coins. Here left and right are adjacent indices of i. After the burst, the left and right then becomes adjacent.

Find the maximum coins you can collect by bursting the balloons wisely.

Note: 
(1) You may imagine nums[-1] = nums[n] = 1. They are not real therefore you can not burst them.
(2) 0 ≤ n ≤ 500, 0 ≤ nums[i] ≤ 100

Example:

Given [3, 1, 5, 8]

Return 167

    nums = [3,1,5,8] --> [3,5,8] -->   [3,8]   -->  [8]  --> []
coins = 3*1*5 + 3*5*8 + 1*3*8 + 1*8*1 = 167

链接: http://leetcode.com/problems/burst-balloons/

题解:

射气球游戏,射中气球以后得分是nums[i] * nums[i - 1] * nums[i - 2],之后左右两边气球相连。 这道题思路也很绕,看了dietpepsi的解答也很模糊,跟买卖股票with cooldown一样。方法应该是用dp或者divide and conquer,大概想法是每burst掉一个气球,我们做一遍2d dp。代码并不长,所以我把答案背下来了...擦。理解交给二刷了。

Time Complexity - O(n3), Space Complexity - O(n2)

public class Solution {
public int maxCoins(int[] orgNums) {
if(orgNums == null || orgNums.length == 0) {
return 0;
}
int len = orgNums.length + 2;
int[] nums = new int[len];
nums[0] = nums[len - 1] = 1; // boundary
for(int i = 0; i < orgNums.length; i++) {
nums[i + 1] = orgNums[i];
}
int[][] dp = new int[len][len]; for(int i = 1; i < len; i++) { // first balloon
for(int lo = 0; lo < len - i; lo++) { // left part
int hi = lo + i; // right part boundary
for(int k = lo + 1; k < hi; k++) {
dp[lo][hi] = Math.max(dp[lo][hi], nums[lo] * nums[k] * nums[hi] + dp[lo][k] + dp[k][hi]);
}
}
} return dp[0][len - 1];
}
}

Reference:

https://leetcode.com/discuss/72216/share-some-analysis-and-explanations

https://leetcode.com/discuss/72186/c-dynamic-programming-o-n-3-32-ms-with-comments

https://leetcode.com/discuss/72215/java-dp-solution-with-detailed-explanation-o-n-3

https://leetcode.com/discuss/72683/my-c-code-dp-o-n-3-20ms

https://leetcode.com/discuss/73288/python-dp-n-3-solutions

https://leetcode.com/discuss/72802/share-my-both-dp-and-divide-conquer-solutions

https://leetcode.com/discuss/73924/my-36ms-c-solution

312. Burst Balloons的更多相关文章

  1. LeetCode 312. Burst Balloons(戳气球)

    参考:LeetCode 312. Burst Balloons(戳气球) java代码如下 class Solution { //参考:https://blog.csdn.net/jmspan/art ...

  2. LN : leetcode 312 Burst Balloons

    lc 312 Burst Balloons 312 Burst Balloons Given n balloons, indexed from 0 to n-1. Each balloon is pa ...

  3. [LeetCode] 312. Burst Balloons 打气球游戏

    Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by ...

  4. 【LeetCode】312. Burst Balloons

    题目: Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented ...

  5. [LeetCode] 312. Burst Balloons 爆气球

    Given n balloons, indexed from 0 to n-1. Each balloon is painted with a number on it represented by ...

  6. 【LeetCode】312. Burst Balloons 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 题目地址: https://leetcode.com/problems/burst-ba ...

  7. 312 Burst Balloons 戳气球

    现有 n 个气球按顺序排成一排,每个气球上标有一个数字,这些数字用数组 nums 表示.现在要求你戳破所有的气球.每当你戳破一个气球 i 时,你可以获得 nums[left] * nums[i] * ...

  8. 312. Burst Balloons - LeetCode

    Question https://leetcode.com/problems/burst-balloons/description/ Solution 题目大意是,有4个气球,每个气球上有个数字,现在 ...

  9. [LeetCode] Burst Balloons (Medium)

    Burst Balloons (Medium) 这题没有做出来. 自己的思路停留在暴力的解法, 时间复杂度很高: 初始化maxCount = 0. 对于当前长度为k的数组nums, 从0到k - 1逐 ...

随机推荐

  1. java 参数化类型

    package com.gxf.collection; import java.util.LinkedList; public class TestForT<T> { private Li ...

  2. c++实例化对象

    今天看到c++实例化对象,有点懵了.Activity_Log the_log (theLogPtr, Tree->GetBranch());这是那一段小代码,开始没看懂.java看习惯了总喜欢n ...

  3. Eclipse中的常用快捷键

    快捷修复 Command+1 //int a=100L; //int a=(int) 100L; 快捷删除行 Command+D 快速起新行 Shift+Enter (当本行代码很长时,将光标定在本行 ...

  4. iOS Automation Test

    google resource for KIF: http://www.oschina.net/translate/ios-ui-testing-with-kif

  5. HttpWatch 安装后在IE上打开

    启动浏览器, 在空白地方左键,  显示出菜单栏 菜单栏中选择"查看">"浏览器栏">"HttpWatch"启动HttpWatch ...

  6. 全面认识JVM技术

    本文向大家描述一下JVM的概念,JVM(Java虚拟机)是可运行Java代码的假想计算机.只要根据JVM规格描述将解释器移植到特定的计算机上,就能保证经过编译的任何Java代码能够在该系统上运行. J ...

  7. input输入框的border-radius属性在IE8下的完美兼容

    在工作中我们发现搜索框大部分都是有圆角的,为此作为经验不足的前端人员很容易就想到,给input标签添加border-radius属性不就解决了嘛.不错方法确实是这样,但是不要忘了border-radi ...

  8. ASP.NET 大文件上传的简单处理

    在 ASP.NET 开发的过程中,文件上传往往使用自带的 FileUpload 控件,可是用过的人都知道,这个控件的局限性十分大,最大的问题就在于上传大文件时让开发者尤为的头疼,而且,上传时无法方便的 ...

  9. Why we have to use epsg:900913 in OpenLayers

    reference:http://docs.openlayers.org/library/spherical_mercator.html epsg:900913 is spicfy the Soher ...

  10. centos mysql 操作

    安装mysqlyum -y install mysql-server 修改mysql配置 vi /etc/my.cnf 这里会有很多需要注意的配置项,后面会有专门的笔记 暂时修改一下编码(添加在密码下 ...