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
Zachery Pole2 小时前
CCF-CSP备战NO.5链表(1)
数据结构·链表
圣光SG4 小时前
Java操作题练习(七)
java·开发语言·算法
Cx330_FCQ5 小时前
Tmux使用
服务器·git·算法
拳里剑气7 小时前
C++算法:多源BFS
c++·算法·宽度优先·多源bfs
ysa0510307 小时前
【板子】短序列dp(换成维护更小常数维度的dp)
c++·笔记·算法·板子
shwill1237 小时前
PID 算法(三)--- 增量 PID ↔ 单神经元 PID 等价映射
linux·算法
wabs6669 小时前
关于图论【卡码网110.字符串迁移的思考】
数据结构·算法·图论
hanlin039 小时前
刷题笔记:力扣第242、349题(哈希表)
笔记·算法·leetcode
aramae9 小时前
C++ IO流完全指南:从C标准库到C++流式编程
服务器·c语言·开发语言·c++·后端
_wyt00110 小时前
c++里的族谱:树
c++·