算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## Range Sum Query 2D - Immutable

### 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.](../images/range-sum-query-2d-immutable.png)

**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**:

* 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.


### 分析

思路跟一维的类似,建立一个累加和矩阵。令状态`f[i][j]`表示从(0,0)到(i,j)的子矩阵的和,则状态转移方程为

`f[i][j] = f[i-1][j] + rowSum`

其中 `rowSum` 是矩阵`matrix[i][0]`到`matrix[i][j]`这一行的和。

有了`f[i][j]`, 则

`sumRange(i1,j1,i2,j2) = f[i2][j2] + f[i1-1][j1-1] - f[i1-1][j2]-f[i2][j1-1]`

将辅助矩阵`f[i][j]`的行数和列数增1,可以简化对矩阵边界的处理。


### 代码

{% if book.java %}
{% codesnippet "./code/range-sum-query-2d-immutable."+book.suffix, language=book.suffix %}{% endcodesnippet %}
{% endif %}


### 相关题目

* [Range Sum Query - Immutable](range-sum-query-immutable.md)
* [Range Sum Query - Mmutable](../binary-tree/segment-tree/range-sum-query-mutable.md)