Title: Range Sum Query 2D – Immutable Source: leetcode.com
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. (Image source: leetcode.com)
Example:
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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 |
Note:
- You may assume that the matrix does not change.
- There are many calls to sumRegion function.
- You may assume that row1 ≤ row2 and col1 ≤ col2.
Python solution
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''' https://leetcode.com/problems/range-sum-query-2d-immutable/ ''' class NumMatrix(object): def __init__(self, matrix): """ initialize your data structure here. :type matrix: List[List[int]] """ rows = len(matrix) # cols = len(matrix[0]) self.sum_mat = [] for i in range(rows): self.sum_mat.append(list(self.accumu(matrix[i]))) # print self.mat # print self.sum_mat def accumu(self, lis): total = 0 for x in lis: total += x yield total def sumRegion(self, row1, col1, row2, col2): """ sum of elements matrix[(row1,col1)..(row2,col2)], inclusive. :type row1: int :type col1: int :type row2: int :type col2: int :rtype: int """ total = 0 for i in range(row1, row2 + 1): total += self.sum_mat[i][col2] # print total, if col1 > 0: total -= self.sum_mat[i][col1-1] # print total, # print total return total |