lintcode :Segmemt Tree Build II
题目
Segmemt Tree Build II
The structure of Segment Tree is a binary tree which each node has two attributes start and end denote an segment / interval.
start and end are both integers, they should be assigned in following rules:
- The root's start and end is given by
buildmethod. - The left child of node A has
start=A.left, end=(A.left + A.right) / 2. - The right child of node A has
start=(A.left + A.right) / 2 + 1, end=A.right. - if start equals to end, there will be no children for this node.
Implement a build method with a given array, so that we can create a corresponding segment tree with every node value represent the corresponding interval max value in the array, return the root of this segment tree.
Given [3,2,1,4]. The segment tree will be:
[0, 3] (max = 4)
/ \
[0, 1] (max = 3) [2, 3] (max = 4)
/ \ / \
[0, 0](max = 3) [1, 1](max = 2)[2, 2](max = 1) [3, 3] (max = 4)
Segment Tree (a.k.a Interval Tree) is an advanced data structure which can support queries like:
- which of these intervals contain a given point
- which of these points are in a given interval
See wiki: Segment Tree Interval Tree
解题
理解题意:根据给的数组构建段树,该节点有区间及其该区间的最大值组成。区间的左右节点利用上面的规则计算。
/**
* Definition of SegmentTreeNode:
* public class SegmentTreeNode {
* public int start, end, max;
* public SegmentTreeNode left, right;
* public SegmentTreeNode(int start, int end, int max) {
* this.start = start;
* this.end = end;
* this.max = max
* this.left = this.right = null;
* }
* }
*/
这个节点定义要好好理解。
/**
* Definition of SegmentTreeNode:
* public class SegmentTreeNode {
* public int start, end, max;
* public SegmentTreeNode left, right;
* public SegmentTreeNode(int start, int end, int max) {
* this.start = start;
* this.end = end;
* this.max = max
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
*@param A: a list of integer
*@return: The root of Segment Tree
*/
public SegmentTreeNode build(int[] A) {
// write your code here
return build(0,A.length-1,A);
}
public SegmentTreeNode build(int start,int end,int[] A){
if(start > end ){
return null;
}
SegmentTreeNode root = new SegmentTreeNode(start,end);
if( start != end){
int mid = (start + end)/2;
root.left = build(start,mid,A);
root.right = build(mid+1,end,A);
root.max = Math.max(root.left.max,root.right.max);
}else{
root.max = A[start];
}
return root;
}
}
Java Code
总耗时: 2532 ms
"""
Definition of SegmentTreeNode:
class SegmentTreeNode:
def __init__(self, start, end, max):
self.start, self.end, self.max = start, end, max
self.left, self.right = None, None
""" class Solution:
# @oaram A: a list of integer
# @return: The root of Segment Tree
def build(self, A):
# write your code here
return self.buildX(0,len(A) - 1,A)
def buildX(self,start,end,A):
if start > end:
return None
maxX = 0
root = SegmentTreeNode(start,end)
if start != end:
mid = int((start + end)/2)
root.left = self.buildX(start,mid,A)
root.right = self.buildX(mid+1,end,A)
root.max = max(root.left.max,root.right.max)
else:
root.max = A[start]
return root
Python Code
总耗时: 750 ms
lintcode :Segmemt Tree Build II的更多相关文章
- [LintCode] Segment Tree Build II 建立线段树之二
The structure of Segment Tree is a binary tree which each node has two attributes startand end denot ...
- lintcode :Coins in Line II 硬币排成线 II
题目 硬币排成线 II 有 n 个不同价值的硬币排成一条线.两个参赛者轮流从左边依次拿走 1 或 2 个硬币,直到没有硬币为止.计算两个人分别拿到的硬币总价值,价值高的人获胜. 请判定 第一个玩家 是 ...
- lintcode:最大子数组II
题目 最大子数组 II 给定一个整数数组,找出两个不重叠子数组使得它们的和最大. 每个子数组的数字在数组中的位置应该是连续的. 返回最大的和. 样例 给出数组[1, 3, -1, 2, -1, 2], ...
- lintcode:Binary Tree Postorder Traversal 二叉树的后序遍历
题目: 二叉树的后序遍历 给出一棵二叉树,返回其节点值的后序遍历. 样例 给出一棵二叉树 {1,#,2,3}, 1 \ 2 / 3 返回 [3,2,1] 挑战 你能使用非递归实现么? 解题: 递归程序 ...
- lintcode :Binary Tree Preorder Traversal 二叉树的前序遍历
题目: 二叉树的前序遍历 给出一棵二叉树,返回其节点值的前序遍历. 样例 给出一棵二叉树 {1,#,2,3}, 1 \ 2 / 3 返回 [1,2,3]. 挑战 你能使用非递归实现么? 解题: 通过递 ...
- 439. Segment Tree Build II
最后更新 08-Jan-2017 开始介绍线段树的主要作用了,可以快速在区间查找极值,我猜是这样的..... 一个NODE的最大值取决于它左边和右边最大值里大 按个,所以,所以什么?对了,我们该用po ...
- Segment Tree Build I & II
Segment Tree Build I The structure of Segment Tree is a binary tree which each node has two attribut ...
- Lintcode247 Segment Tree Query II solution 题解
[题目描述] For an array, we can build a Segment Tree for it, each node stores an extra attribute count t ...
- [Locked] Closest Binary Search Tree Value & Closest Binary Search Tree Value II
Closest Binary Search Tree Value Given a non-empty binary search tree and a target value, find the ...
随机推荐
- jQuery Mobile 开发中常见的问题
目录 页面缩放显示问题 页面跳转后样式丢失js失效 跳转时重复调用pageinit方法的解决办法 如何调用loading效果 1页面缩放显示问题 问题描述: 页面似乎被缩小了,屏幕太宽了. 解决办 ...
- ASP.NET MVC 及 Areas 简单控制路由
ASP.NET MVC中怎么去控制路由,这个想关的文章很多,我在这里就是自我总结一下,仅供参考. 1.我们新建一个项目,查看RouteConfig.cs,代码如下: public static voi ...
- 转:java中volatile关键字的含义
转:java中volatile关键字的含义 在java线程并发处理中,有一个关键字volatile的使用目前存在很大的混淆,以为使用这个关键字,在进行多线程并发处理的时候就可以万事大吉. Java语言 ...
- Oracle表空间、段、区和块
数据块(Block) 数据块Block是Oracle存储数据信息的最小单位.注意,这里说的是Oracle环境下的最小单位.Oracle也就是通过数据块来屏蔽不同操作系统存储结构的差异.无论是Windo ...
- objdump的使用方法和 symbol table的每列的含义
一.objdump的用法 objdump命令的man手册 objdump [-a] [-b bfname| --target=bfdname] [-C] [--debugging] ...
- django post分号引发的问题
利用jquery的ajax传值 $.ajax({ type:"POST", url:"", data:"content"=content, ...
- eclipse安装pydev插件
打开Eclipse,找到Help菜单栏,进入Install New Software…选项. 点击work with:输入框的旁边点击Add…,Name可以随便输入,Location是http://p ...
- Hadoop命令摘录
一:文件操作 1.建立目录 [hadoop@hadoop1:hadoop]$bin/hadoop dfs -mkdir testdir 在HDFS中建立一个名为testdir的目录 2.上传文件到HD ...
- UNIX环境高级编程-环境配置
环境配置步骤如下. 1. 下载源文件:http://www.apuebook.com/src.tar.gz. 2. 复制src.tar.gz文件到/home/me/mydir/unixl/目录(自 ...
- Java Day 05
数组第二种定义 数组-遍历 数组操作的核心思想就是对角标的操作: 数组-求最值 1.循环 比较 排序 选择排序 把原始数组分割成了两个数组,至少有一个是有序的 冒泡排序 相邻元素比较 位置置换代码提取 ...