Range Sum Query 2D

Range Sum Query 2D - Immutable

题目来源：https://leetcode.com/problemset/algorithms/
题目类型：动态规划

-题目描述-
-解题思路-
-代码实现-

题目描述

Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).

Range Sum Query 2D
The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.

Example:
Given matrix = [
[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]
]
sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12

解题思路

题意为计算一个矩阵中指定矩形内的元素的和。即将matrix[i][j](row1<=i<= row2&&col1<=j<= col2)累加就可以得到结果。

代码实现

@requires_authorizationclass NumMatrix {public:    NumMatrix(vector<vector<int>> matrix) {        for (int i = 0; i < matrix.size(); i++) {            vector<int> col;            for (int j = 0; j < matrix[i].size(); j++) {                col.push_back(matrix[i][j]);            }            this->matrix.push_back(col);        }    }    int sumRegion(int row1, int col1, int row2, int col2) {        if (!(row1 <= row2&&col1 <= col2)) return 0;        int sum = 0;        for (int i = row1; i <= row2; i++) {            for (int j = col1; j <= col2; j++) {                sum += matrix[i][j];            }        }        return sum;    }    private:    vector<vector<int>> matrix;};