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
倒头就睡的小比特1 天前
算法竞赛C++常用的STL
c++·算法
weilx12341 天前
C++笔记-文件IO-<fcntl.h>
c++
小羊没烦恼!1 天前
初探性能优化——2个月到4小时的性能提升
java·开发语言·windows·算法·c#
猎头南楼1 天前
知识社区推荐系统实践:新用户冷启动与长短期兴趣建模的挑战 资深推荐算法工程师
人工智能·深度学习·算法·机器学习
Smileyqp沛沛1 天前
前端?C++ ?较大差异基础罗列
c++·基础·前端转c++
m0_547486661 天前
《数据结构教程》全套 PPT课件2026
数据结构
C语言小火车1 天前
C/C++ 为什么需要编译器?
开发语言·c++
旖旎夜光1 天前
力控面试题 01.01: 判定字符是否唯一(位运算) —— 题解
c++·学习·算法·leetcode·力控
wzdark1 天前
大规模并行计算中的负载均衡算法研究4
算法
吞下星星的少年·-·1 天前
C++ 萌新语法入门篇
c++·算法比赛