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
得物技术1 小时前
从狂野代码到按目标生产:得物推荐 AI Harness 的工程化实践|AICon 演讲整理
人工智能·算法·架构
AI小老六4 小时前
SkillOpt 架构拆解:把 Skill 文本当参数,用执行轨迹训练 Agent
后端·算法·ai编程
胡萝卜术5 小时前
从“分数打架”到“排名投票”:为什么你的ChatBI必须用RRF?
算法·设计模式·面试
Asize5 小时前
初识DFS 与 BFS:递归、队列与图遍历
算法
罗西的思考19 小时前
机器人 / 强化学习】HIL-SERL:人类在环驱动的具身智能进化框架
人工智能·算法·机器学习
CSharp精选营21 小时前
关系型 vs 非关系型:从原理到选型,一文搞定数据库核心分类
数据结构·nosql·关系型数据库·非关系型数据库·技术选型
美团技术团队1 天前
LongCat 开源 VitaBench 2.0:长期动态智能体基准新标杆
人工智能·算法
用户805533698031 天前
不止三件套:QObject 属性系统全关键字与运行时反射!
c++·qt
To_OC2 天前
LC 207 课程表:刚学图论那会儿,我连这是拓扑排序都没看出来
javascript·算法·leetcode
To_OC2 天前
LC 208 实现 Trie 前缀树:曾被名字劝退,写完发现是送分题
javascript·算法·leetcode