Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. Each query has two integers [start, end]. For each query, calculate the sum number between index start and end in the given array, return the result list.

Example

For array [1,2,7,8,5], and queries [(0,4),(1,2),(2,4)], return [23,9,20].

Analysis:
Use an array to save the sum from 0 to i. Then for query [i, j], we shoud return sum[j] - sum[i - 1].

 /**
* Definition of Interval:
* public classs Interval {
* int start, end;
* Interval(int start, int end) {
* this.start = start;
* this.end = end;
* }
*/
public class Solution {
/**
*@param A, queries: Given an integer array and an query list
*@return: The result list
*/
public ArrayList<Long> intervalSum(int[] A,
ArrayList<Interval> queries) { ArrayList<Long> list = new ArrayList<Long>();
if (A == null || A.length == ) return list;
if (queries == null || queries.size() == ) return list; long[] sum = new long[A.length]; for (int i = ; i < sum.length; i++) {
if (i == ) {
sum[i] = A[i];
} else {
sum[i] += sum[i - ] + A[i];
}
} for (int i = ; i < queries.size(); i++) {
Interval interval = queries.get(i);
if (interval.start == ) {
list.add(sum[interval.end]);
} else {
list.add(sum[interval.end] - sum[interval.start - ]);
}
}
return list;
}
}

Interval Sum II

Given an integer array in the construct method, implement two methods query(start, end)and modify(index, value):

  • For query(startend), return the sum from index start to index end in the given array.
  • For modify(indexvalue), modify the number in the given index to value
Example

Given array A = [1,2,7,8,5].

  • query(0, 2), return 10.
  • modify(0, 4), change A[0] from 1 to 4.
  • query(0, 1), return 6.
  • modify(2, 1), change A[2] from 7 to 1.
  • query(2, 4), return 14.

Analysis:

As the value in the array may change, so it is better to build a segement tree. If the value in the tree is changed, we also need to update its parent node.

 public class Solution {
/* you may need to use some attributes here */ SegmentTreeNode root; /**
* @param A:
* An integer array
*/
public Solution(int[] A) {
if (A == null || A.length == )
return;
root = build(A, , A.length - );
} private SegmentTreeNode build(int[] A, int start, int end) {
if (A == null || start < || end >= A.length)
return null;
SegmentTreeNode root = new SegmentTreeNode(start, end, );
if (start == end) {
root.sum = A[start];
} else {
int mid = (end - start) / + start;
root.left = build(A, start, mid);
root.right = build(A, mid + , end);
root.sum = root.left.sum + root.right.sum;
}
return root;
} public long query(int start, int end) {
if (root == null || start > end)
return ;
return helper(root, Math.max(, start), Math.min(end, root.end));
} public long helper(SegmentTreeNode root, int start, int end) {
if (root.start == start && root.end == end) {
return root.sum;
} int mid = (root.start + root.end) / ;
if (start >= mid + ) {
return helper(root.right, start, end);
} else if (end <= mid) {
return helper(root.left, start, end);
} else {
return helper(root.left, start, mid) + helper(root.right, mid + , end);
}
} public void modify(int index, int value) {
if (root == null)
return;
if (index < root.start && index > root.end)
return;
modifyHelper(root, index, value);
} public void modifyHelper(SegmentTreeNode root, int index, int value) {
if (root.start == root.end && root.start == index) {
root.sum = value;
return;
} int mid = (root.start + root.end) / ;
if (index >= mid + ) {
modifyHelper(root.right, index, value);
} else {
modifyHelper(root.left, index, value);
}
root.sum = root.left.sum + root.right.sum; }
} class SegmentTreeNode {
public int start, end, sum;
public SegmentTreeNode left, right; public SegmentTreeNode(int start, int end, int sum) {
this.start = start;
this.end = end;
this.sum = sum;
this.left = this.right = null;
}
}

Interval Sum I && II的更多相关文章

  1. Lintcode: Interval Sum II

    Given an integer array in the construct method, implement two methods query(start, end) and modify(i ...

  2. 1. Two Sum I & II & III

    1. Given an array of integers, return indices of the two numbers such that they add up to a specific ...

  3. [Leetcode][JAVA] Path Sum I && II

    Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that addi ...

  4. LeetCode:Combination Sum I II

    Combination Sum Given a set of candidate numbers (C) and a target number (T), find all unique combin ...

  5. LeetCode:Path Sum I II

    LeetCode:Path Sum Given a binary tree and a sum, determine if the tree has a root-to-leaf path such ...

  6. Nested List Weight Sum I & II

    Nested List Weight Sum I Given a nested list of integers, return the sum of all integers in the list ...

  7. Two Sum I & II

    Two Sum I Given an array of integers, find two numbers such that they add up to a specific target nu ...

  8. Lintcode: Interval Sum

    Given an integer array (index from 0 to n-1, where n is the size of this array), and an query list. ...

  9. Leetcode 39 40 216 Combination Sum I II III

    Combination Sum Given a set of candidate numbers (C) and a target number (T), find all unique combin ...

随机推荐

  1. java实验报告一

    一.实验内容 1. 使用JDK编译.运行简单的Java程序 2.使用Eclipse 编辑.编译.运行.调试Java程序 二.实验步骤 (一)命令行下Java程序开发 1. 首先双击桌面上的Xface终 ...

  2. 《Linux内核设计与实现》Chapter 5 读书笔记

    <Linux内核设计与实现>Chapter 5 读书笔记 在现代操作系统中,内核提供了用户进程与内核进行交互的一组接口,这些接口的作用是: 使应用程序受限地访问硬件设备 提供创建新进程与已 ...

  3. asp.net 网页拉伸 到300%不变形方法一

    网页拉伸到300%控件和表格不会出现太大变形 方法: 1.对主页面采用百分比宽度(Width="100%") 2.对于表格使用百分比宽度,包括表格宽度和表格中顶端td宽度 3.对t ...

  4. linux 远程连接报错 10038或者10061 或者10060

    1.检查linux的mysql是否开启 2.检查mysql的user表的host是否是% 3.检查my.cnf文件是否绑定本地 4.防火墙3306端口是否开启 假如以上都没问题,那最大的原因就是我折腾 ...

  5. [福大软工] Z班 团队作业——UML设计 作业成绩

    团队作业--UML设计 作业链接 http://www.cnblogs.com/easteast/p/7745703.html 作业要求 1)团队分工(5分) 描述团队的每个成员分别完成了UML图的哪 ...

  6. 团队作业四-WBS练习

    我们团队开发的是四则运算,主要面对的用户是小学生.老师及学生家长.经过我们组成员的讨论和结合实际及自身能力,对团队成员分配任务,队长负责全局工作主要负责任务,统一进度,和适量的编码,露哥和阮磊主要负责 ...

  7. DEP

    DEP(Data execution protect)数据执行保护,这个功能需要操作系统和硬件的共同支持才可以生效.DEP的原理就是在系统的内存页中设置了一个标志位,标示这个内存页的属性(可执行). ...

  8. 『编程题全队』Alpha阶段发布说明

    1. 这一版本的功能 (1)管理个人的任务事项,管理用户的提醒事项,提供一个简洁的操作界面,将其分类为全部.今天.明日.最近七天.更远.还有已完成,方便用户进行事务管理和整理. (2)提供一个便捷的备 ...

  9. jmeter测试soap协议时候 路径不需要添加

  10. BZOJ4559 JLOI2016成绩比较(容斥原理+组合数学+斯特林数)

    容斥一发改为计算至少碾压k人的情况数量,这样对于每门课就可以分开考虑再相乘了.剩下的问题是给出某人的排名和分数的值域,求方案数.枚举出现了几种不同的分数,再枚举被给出的人的分数排第几,算一个类似斯特林 ...