Algorithm

每日一刷-Range-Sum-Query-2D-Immutable

[每日一刷] (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:
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


这题思路和一维的时候一致，可以使用动态规划也可以使用线段树。就这题来说使用动态规划会简单很多。
思路就是用dp[i][j]表示从(0, 0)到(i, j)的和

Python:

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
#!/usr/bin/env 
# -*- coding:utf-8 -*-
"""
时间复杂度 Query O(1), Build O(n^2)
"""
class NumMatrix(object):
    def __init__(self, matrix):
        """
        initialize your data structure here.
        :type matrix: List[List[int]]
        """
        if matrix == []:
            return
        r = len(matrix)
        c = len(matrix[0])
        self.dp = [[0 for i in xrange(c+1)] for j in xrange(r+1)]
        for i in xrange(1, r+1):
            for j in xrange(1, c+1):
                self.dp[i][j] = self.dp[i-1][j] + self.dp[i][j-1] - self.dp[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.dp[row2+1][col2+1] - self.dp[row1][col2+1] - self.dp[row2+1][col1] + self.dp[row1][col1]