Range Sum Query - Immutable

Given an integer array nums, find the sum of the elements between indices i and j (i ≤ j), inclusive.

Example:

Given nums = [-2, 0, 3, -5, 2, -1]

sumRange(0, 2) -> 1
sumRange(2, 5) -> -1
sumRange(0, 5) -> -3

Note:

  1. You may assume that the array does not change.
  2. There are many calls to sumRange function.
 class NumArray {
private:
vector<int> acc; public:
NumArray(vector<int> &nums) {
acc.push_back();
for (auto n : nums) {
acc.push_back(acc.back() + n);
}
} int sumRange(int i, int j) {
return acc[j + ] - acc[i];
}
}; // Your NumArray object will be instantiated and called as such:
// NumArray numArray(nums);
// numArray.sumRange(0, 1);
// numArray.sumRange(1, 2);

Range Sum Query 2D - Immutable

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).


The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

Example:

Given matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
] sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12

Note:

  1. You may assume that the matrix does not change.
  2. There are many calls to sumRegion function.
  3. You may assume that row1 ≤ row2 and col1 ≤ col2.
 class NumMatrix {
private:
vector<vector<int>> acc;
public:
NumMatrix(vector<vector<int>> &matrix) {
if (matrix.empty()) return;
int n = matrix.size(), m = matrix[].size();
acc.resize(n + , vector<int>(m + ));
for (int i = ; i <= n; ++i) acc[i][] = ;
for (int j = ; j <= m; ++j) acc[][j] = ;
for (int i = ; i <= n; ++i) {
for (int j = ; j <= m; ++j) {
acc[i][j] = acc[i][j-] + acc[i-][j] - acc[i-][j-] + matrix[i-][j-];
}
}
} int sumRegion(int row1, int col1, int row2, int col2) {
return acc[row2+][col2+] - acc[row1][col2+] - acc[row2+][col1] + acc[row1][col1];
}
}; // Your NumMatrix object will be instantiated and called as such:
// NumMatrix numMatrix(matrix);
// numMatrix.sumRegion(0, 1, 2, 3);
// numMatrix.sumRegion(1, 2, 3, 4);

[LeetCode] Range Sum Query - Immutable & Range Sum Query 2D - Immutable的更多相关文章

  1. [LeetCode] Range Sum Query 2D - Immutable 二维区域和检索 - 不可变

    Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper lef ...

  2. 【LeetCode】304. Range Sum Query 2D - Immutable 解题报告(Python)

    作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 预先求和 相似题目 参考资料 日期 题目地址:htt ...

  3. 【刷题-LeetCode】304. Range Sum Query 2D - Immutable

    Range Sum Query 2D - Immutable Given a 2D matrix matrix, find the sum of the elements inside the rec ...

  4. LeetCode OJ:Range Sum Query 2D - Immutable(区域和2D版本)

    Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper lef ...

  5. LeetCode 304. Range Sum Query 2D – Immutable

    Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper lef ...

  6. 304. Range Sum Query 2D - Immutable

    题目: Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper ...

  7. LeetCode-304. Range Sum Query 2D - Immutable

    Description: Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by ...

  8. LeetCode 548. Split Array with Equal Sum (分割数组使得子数组的和都相同)$

    Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies fol ...

  9. [LeetCode] 548. Split Array with Equal Sum 分割数组成和相同的子数组

    Given an array with n integers, you need to find if there are triplets (i, j, k) which satisfies fol ...

随机推荐

  1. Hibernate映射问题之OneToOne【自己整理】

    首先贴上一个MkYong的例子 stock.java package com.mkyong.stock; import javax.persistence.CascadeType; import ja ...

  2. 黑马程序员——for循环的使用与理解

    Console.WriteLine("请输入要打印菱形的行数(不能是偶数)");---------------------- <a href="http://edu ...

  3. C#设计模式(11)——外观模式(Facade Pattern)

    一.引言 在软件开发过程中,客户端程序经常会与复杂系统的内部子系统进行耦合,从而导致客户端程序随着子系统的变化而变化,然而为了将复杂系统的内部子系统与客户端之间的依赖解耦,从而就有了外观模式,也称作 ...

  4. WebApi 服务监控

    本文主要介绍在请求WebApi时,监控Action执行的时间,及Action传递的参数值,以及Http请求头信息.采用log4net记录监控日志,通过日志记录的时间方便我们定位哪一个Action执行的 ...

  5. 深入理解java虚拟机【Java虚拟机垃圾收集器】

    Java堆内存被划分为新生代和年老代两部分,新生代主要使用复制和标记-清除垃圾回收算法,年老代主要使用标记-整理垃圾回收算法,因此java虚拟中针对新生代和年老代分别提供了多种不同的垃圾收集器,JDK ...

  6. [WinAPI] API 7 [判断光驱内是否有光盘]

    判断光驱中是否有光盘,仍然可以使用GetDriveType和GetVolumeInformation函数实现.首先使用驱动器根路径作为GetDriveType和参数,如果返回值是DRIVE_CDROM ...

  7. [JS10] 获取时间

    <html> <head> <meta http-equiv="Content-Type" content="text/html; char ...

  8. 通过实验窥探javascript的解析执行顺序

    简介 javascript是一种解释型语言,它的执行是自上而下的.但是各浏览器对于[自上而下]的理解是有细微差别的,而代码的上下游也就是程序流对于程序正确运行又是至关重要的.所以我们有必要深入理解js ...

  9. canvas/CSS实现仪表盘效果

    手机上看比较虚 <!DOCTYPE html> <html> <head> <meta charset="UTF-8"> <t ...

  10. SQL语句的基础

    注释语法:#注释语 一.T-SQL语句注意:1.语句写完后用"分号:"代表这一句结束2.列结束用逗号,最后一列写完不用写逗号3.符号一定是英文的 关键字:主键:primary ke ...