# LeetCode Range Sum Query 2D - Immutable

LeetCode 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).

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.

dp[i][j]=dp[i][j-1]+dp[i-1][j]-dp[i-1][j-1]+matrix[i][j]

cursum=dp[j][v]-dp[j][u-1]-dp[i-1][v]+dp[i-1][u-1]

```      |               |
-----(i,u)----------(i,v)----
|               |
|               |
-----(j,u)----------(j,v)----
|               |```

```class NumMatrix {
private:
vector<vector<int>> dp;
public:
NumMatrix(vector<vector<int>> matrix) {
if (matrix.empty() || matrix[0].empty())return;
int m = matrix.size(), n = matrix[0].size();
for (int i = 0; i < m + 1; ++i)dp.push_back(vector<int>(n + 1, 0)); // 为了方便，多增加了一行和一列
for (int i = 1; i <= m; ++i) {
for (int j = 1; j <= n; ++j) {
dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1] + matrix[i - 1][j - 1];
}
}
}

int sumRegion(int row1, int col1, int row2, int col2) {
if (dp.empty())return 0;
++row1; // 因为dp多一行和一列，所以这里提前加上
++col1;
++row2;
++col2;
return dp[row2][col2] - dp[row2][col1 - 1] - dp[row1 - 1][col2] + dp[row1 - 1][col1 - 1];
}
};
```