[LeetCode] Range Sum Query - Immutable & Range Sum Query 2D - 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:
- You may assume that the array does not change.
- 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:
- You may assume that the matrix does not change.
- There are many calls to sumRegion function.
- 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
- [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 ...
- 【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 ...
- 【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 ...
- 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 ...
- 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 ...
- 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 ...
- LeetCode-304. Range Sum Query 2D - Immutable
Description: Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by ...
- 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 ...
- [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
- CCF 201612-1 最大波动 (水题)
问题描述 小明正在利用股票的波动程度来研究股票.小明拿到了一只股票每天收盘时的价格,他想知道,这只股票连续几天的最大波动值是多少,即在这几天中某天收盘价格与前一天收盘价格之差的绝对值最大是多少. 输入 ...
- MySQL大小写敏感说明
Linux环境下,不是windows平台下.区别很大.注意. 一图胜千言 mysql> show create table Ac; +-------+-------------------- ...
- GridView获取当前行
int row = ((GridViewRow)((DropDownList)sender).NamingContainer).RowIndex; //获取GridView里的DropDownList ...
- Duang的成长——使用造字程序输入生僻字
使用造字程序输入生僻字 最近,一个字突然间火了起来,那就是——duang! (图片来自网络) 那么,问题来了!造字程序哪家强?(此处有掌声) 其实,微软早就考虑到各国文字的博大精深,在系统中集成了一个 ...
- Q114寒假作业之割绳子
割绳子 TimeLimit:1000MS MemoryLimit:10000K 64-bit integer IO format:%lld Problem Description 已知有n条绳子,每 ...
- SQLSERVER 数据库性能的的基本
SQLSERVER 数据库性能的基本 很久没有写文章了,在系统正式上线之前,DBA一般都要测试一下服务器的性能 比如你有很多的服务器,有些做web服务器,有些做缓存服务器,有些做文件服务器,有些做数据 ...
- Dynamic CRM 2013学习笔记(二十)字段改变事件的二种实现方法
CRM里有二种方式实现字段change事件,一种是在form里,一种完全通过js来实现.本文介绍下二者的用途及区别. 1. Form里用法 这种方式估计其实也是添加一个js的function. 这种方 ...
- 在Eclipse添加Android兼容包( v4、v7 appcompat )
昨天添加Android兼容包,碰到了很多问题,在这里记录一下,让后面的路好走. 如何选择兼容包, 请参考Android Support Library Features(二) 一.下载Support ...
- [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 ...
- 使用Nito.AsyncEx实现异步锁
Lock是常用的同步锁,但是我们无法在Lock的内部实现异步调用,比如我们无法使用await. 以下面的代码为例,当你在lock内部使用await时,VS会报错提醒. 最简单的解决办法就是使用第三方的 ...