LeetCode 题解(287): Range Sum Query 2D - Immutable

题目：

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.

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

Note:

1. You may assume that the matrix does not change.
2. There are many calls to sumRegion function.
3. You may assume that row1 ≤ row2 and col1 ≤ col2.

题解：

和上题一样，只不过变成二维。

Python版：

class NumMatrix(object):
    def __init__(self, matrix):
        """
        initialize your data structure here.
        :type matrix: List[List[int]]
        """
        self.sums = [[0 for i in range(len(matrix[0]))] for j in range(len(matrix))]
        for i in range(len(matrix)):
            for j in range(len(matrix[0])):
                self.sums[i][j] = matrix[i][j] if j == 0 else self.sums[i][j-1] + matrix[i][j]

    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
        """
        sums = 0
        for i in range(row1, row2+1):
            sums += self.sums[i][col2] - self.sums[i][col1 - 1] if col1 != 0 else self.sums[i][col2];
        return sums


# Your NumMatrix object will be instantiated and called as such:
# numMatrix = NumMatrix(matrix)
# numMatrix.sumRegion(0, 1, 2, 3)
# numMatrix.sumRegion(1, 2, 3, 4)