卖萌的弱渣

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Range Sum Query 2D - Immutable

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).

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.

Solution

  • sums[i][j]: nums[0][0]+…+nums[i][j]
(Range-Sum-Query-2D-Immutable.py) download
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class NumMatrix(object):
    def __init__(self, matrix):
        """
        initialize your data structure here.
        :type matrix: List[List[int]]
        """
        if not matrix:
            return
        row = len(matrix)
        col = len(matrix[0])
        self.sums = [[0]*(col+1) for x in range(row+1)]
        for i in range(1,row+1):
            for j in range(1,col+1):
                self.sums[i][j] = self.sums[i-1][j] + self.sums[i][j-1] - self.sums[i-1][j-1] + matrix[i-1][j-1]

    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
        """
        return self.sums[row2+1][col2+1]-self.sums[row2+1][col1]+self.sums[row1][col1]-self.sums[row1][col2+1]


# 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)