Question
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).
The above rectangle (with the red border) is defined by (row1, col1) = (2, 1) and (row2, col2) = (4, 3), which contains sum = 8.
You may assume that the matrix does not change.You may assume that row1 ≤ row2 and col1 ≤ col2. There are many calls to sumRegion function.
给出一个二维矩阵以及左上角坐标(row1, col1)和右上角坐标(row2, col2),求出以它们框出的矩形中的元素的总和
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
Solution
动态规划解。求得上图的红框矩形其实我们可以在sum_matrix[i][j]中保存matrix[0][0]-matrix[i][j]的元素总和,然后如下图
(r1, c1) - (r2, c2)(红框) = sum_matrix[r2][c2] - sum_matrix[r2][c1 - 1](黄框) - sum_matrix[r1 - 1][c2](粉框) + sum_matrix[r1 - 1][c1 - 1](黄框和粉框重合部分)
那么如何得到sum_matrix呢,其实是一样的思路,在遍历矩阵matrix时,递推式为:sum_matrix[i][j] = sum_matrix[i - 1][j] + sum_matrix[i][j - 1] - sum_matrix[i - 1][j - 1] + matrix[i][j]
class NumMatrix(object): def __init__(self, matrix): """ :type matrix: List[List[int]] """ # matrix = [],不符合二维矩阵 if len(matrix) == 0: return self.sum_matrix = [[0 for _ in range(len(matrix[0]) + 1)] for _ in range(len(matrix) + 1)] print(self.sum_matrix) for index_r in range(1, len(matrix) + 1): for index_c in range(1, len(matrix[0]) + 1): self.sum_matrix[index_r][index_c] = self.sum_matrix[index_r - 1][index_c] + \ self.sum_matrix[index_r][index_c - 1] - \ self.sum_matrix[index_r - 1][index_c - 1] + \ matrix[index_r - 1][index_c - 1] def sumRegion(self, row1, col1, row2, col2): """ :type row1: int :type col1: int :type row2: int :type col2: int :rtype: int """ # 因为sum_matrix的矩阵和matrix有所不同,所以坐标全部要+1 return self.sum_matrix[row2 + 1][col2 + 1] - self.sum_matrix[row2 + 1][col1] - \ self.sum_matrix[row1][col2 + 1] + self.sum_matrix[row1][col1]
本文介绍了一种使用动态规划高效解决二维矩阵区间和查询的方法。通过预处理计算累积和矩阵,实现快速响应多次区间和查询请求。适用于图像处理、数据分析等场景。
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