Range Sum Query 2D - Immutable

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:

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.

解题思路：
类似于题目Range Sum Query - Immutable的解题思路。使用一个二维辅助数组m_regionSum保存matrix[0:i][0:j]区域的元素的sum。

m_regionSum元素的更新方法为：m_regionSum[i][j]=m_regionSum[i-1][j]+m_regionSum[i][j-1]+m_regionSum[i-1][j-1]+matrix[i][j]

则sumRegion(row1,col1,row2,col2)=m_regionSum[row2][col2]-m_regionSum[row1-1][col2]-m_regionSum[row2][col1-1]+m_regionSum[row-1][col1-1]

AC代码如下：

class NumMatrix {
public:
NumMatrix(vector<vector<int>> &matrix) {
if (matrix.size() > 0 && matrix[0].size() > 0){
int n = matrix.size();
int m = matrix[0].size();
m_regionSum = vector<vector<int>>(n+1, vector<int>(m+1, 0));
for (int i = 1; i <= n; ++i){
for (int j = 1; j <= m; ++j){
m_regionSum[i][j] = m_regionSum[i - 1][j] + m_regionSum[i][j - 1] -
m_regionSum[i - 1][j - 1] + matrix[i - 1][j - 1];
}
}
}
}

int sumRegion(int row1, int col1, int row2, int col2) {
return m_regionSum[row2 + 1][col2 + 1] - m_regionSum[row1][col2 + 1] -
m_regionSum[row2 + 1][col1] + m_regionSum[row1][col1];
}
private:
vector<vector<int>> m_regionSum;
};