# 304. Range Sum Query 2D - Immutable 二维区域和检索 - 矩阵不可变

@TOC

## # 题目描述

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.

## # 解题方法

### # 预先求和

Sum(ABCD)=Sum(OD)−Sum(OB)−Sum(OC)+Sum(OA)

class NumMatrix(object):

def __init__(self, matrix):
"""
:type matrix: List[List[int]]
"""
if not matrix or not matrix[0]:
M, N = 0, 0
else:
M, N = len(matrix), len(matrix[0])
self.sumM = [[0] * (N + 1) for _ in range(M + 1)]
for i in range(M):
for j in range(N):
self.sumM[i + 1][j + 1] = self.sumM[i][j + 1] + self.sumM[i + 1][j]  - self.sumM[i][j] + matrix[i][j]

def sumRegion(self, row1, col1, row2, col2):
"""
:type row1: int
:type col1: int
:type row2: int
:type col2: int
:rtype: int
"""
return self.sumM[row2 + 1][col2 + 1] - self.sumM[row2 + 1][col1] - self.sumM[row1][col2 + 1] + self.sumM[row1][col1]

# Your NumMatrix object will be instantiated and called as such:
# obj = NumMatrix(matrix)
# param_1 = obj.sumRegion(row1,col1,row2,col2)

## # 相似题目

303. Range Sum Query - Immutableopen in new window

## # 参考资料

https://leetcode.com/articles/range-sum-query-2d-immutable/

## # 日期

2018 年 10 月 30 日 —— 啊，十月过完了