LeetCode304. Range Sum Query 2D - Immutable

文章目录

一、题目

Given a 2D matrix matrix, handle multiple queries of the following type:

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

Implement the NumMatrix class:

NumMatrix(int\[\]\[\] matrix) Initializes the object with the integer matrix matrix.

int sumRegion(int row1, int col1, int row2, int col2) Returns the sum of the elements of matrix inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

You must design an algorithm where sumRegion works on O(1) time complexity.

Example 1:

Input

"NumMatrix", "sumRegion", "sumRegion", "sumRegion"

\[\[\[3, 0, 1, 4, 2\], \[5, 6, 3, 2, 1\], \[1, 2, 0, 1, 5\], \[4, 1, 0, 1, 7\], \[1, 0, 3, 0, 5\]\]\], \[2, 1, 4, 3\], \[1, 1, 2, 2\], \[1, 2, 2, 4\]

Output

null, 8, 11, 12

Explanation

NumMatrix numMatrix = new NumMatrix(\[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]);

numMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)

numMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)

numMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)

Constraints:

m == matrix.length

n == matrixi.length

1 <= m, n <= 200

-104 <= matrixij <= 104

0 <= row1 <= row2 < m

0 <= col1 <= col2 < n

At most 104 calls will be made to sumRegion.

二、题解

cpp 复制代码
class NumMatrix {
public:
    int sum[205][205];
    NumMatrix(vector<vector<int>>& matrix) {
        int m = matrix.size();
        int n = matrix[0].size();
        for(int i = 0;i <= m;i++) sum[i][0] = 0;
        for(int j = 0;j <= n;j++) sum[0][j] = 0;
        for(int i = 1;i < m + 1;i++){
            for(int j = 1;j < n + 1;j++){
                sum[i][j] = matrix[i-1][j-1];
            }
        }
        for(int i = 1;i <= m;i++){
            for(int j = 1;j <= n;j++){
                sum[i][j] += sum[i][j-1] + sum[i-1][j] - sum[i-1][j-1];
            }
        }
    }
    
    int sumRegion(int row1, int col1, int row2, int col2) {
        row2++;
        col2++;
        return sum[row2][col2] - sum[row2][col1] - sum[row1][col2] + sum[row1][col1];
    }
};

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix* obj = new NumMatrix(matrix);
 * int param_1 = obj->sumRegion(row1,col1,row2,col2);
 */
相关推荐
手写码匠14 小时前
手写 GraphRAG:从零实现图增强检索增强生成系统
人工智能·深度学习·算法·aigc
BomanGe114 小时前
NSK重载高刚性滚珠丝杠技术详解
经验分享·算法·规格说明书
赴生-14 小时前
C++进阶 C++11(下)
开发语言·c++
有点。15 小时前
C++(贪心算法一)
c++·贪心算法
Matrix_1115 小时前
手机里的计算摄影:广角形变校正算法
人工智能·算法·智能手机·计算摄影
WBluuue15 小时前
数据结构与算法:有序表(二):跳表
数据结构·c++·算法·skiplist
赴生-15 小时前
C++进阶 异常
开发语言·c++
不好听61316 小时前
深入理解链表:线性数据结构的另一面
javascript·数据结构
x1387028595716 小时前
c语言中srtlen(指针使用计算字符长度)、传值和传址调用
c语言·开发语言·算法·visual studio
海兰16 小时前
【实用程序】电商销售分析仪表盘 — 从零搭建一个AI参与的全栈数据洞察系统
人工智能·学习·算法