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
生态学者9 小时前
Global Change Biology | 森林越“步调一致”,反而越怕干旱?树木生长与叶片活动的同步之谜
大数据·人工智能·算法·r语言·微信公众平台
所以遗憾是什么呢?9 小时前
【题解】Educational Codeforces Round 195 (Rated for Div. 2) ABCDE
c++·动态规划·acm·icpc·ccpc·xcpc·codeforecs
lsswear10 小时前
Agent 框架新手入门与实战部署指南
大数据·人工智能·算法
学逆向的10 小时前
C 循环结构的执行效率分析
c语言·开发语言·数据结构·网络安全
Lhan.zzZ11 小时前
Qt WebEngineView 在 Linux 上无法加载网页:从“看不见 HTML“到“禁用沙箱“
linux·c++·qt
Data analyse45611 小时前
rnweb 回放数据量太大怎么办?分片存储与断点续传的实现思路
前端·网络·算法·数据分析
sysu_chenxy12 小时前
二重循环输出二维图形
c++
Navigator_Z13 小时前
LeetCode //C - 1293. Shortest Path in a Grid with Obstacles Elimination
c语言·算法·leetcode
Nil20813 小时前
leetcode 64最小路径和
算法·leetcode·职场和发展
光影少年13 小时前
AI Agent、自动化与未来工作
人工智能·算法·架构