## 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).

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

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

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

### 代码

// Range Sum Query 2D - Immutable
public class NumMatrix {
// Time Complexity: O(n*m), Space Complexity: O(1)
public NumMatrix(int[][] matrix) {
final int m = matrix.length;
final int n = matrix.length > 0 ? matrix[0].length : 0;
this.f = new int[m + 1][n + 1];

for (int i = 1; i < m + 1; ++i) {
int rowSum = 0;
for (int j = 1; j < n + 1; ++j) {
f[i][j] += rowSum + matrix[i-1][j-1];
if (i > 1) {
f[i][j] += f[i-1][j];
}
rowSum += matrix[i-1][j-1];
}
}
}

// Time Complexity: O(1), Space Complexity: O(1)
public int sumRegion(int row1, int col1, int row2, int col2) {
return f[row2 + 1][col2 + 1] + f[row1][col1] -
f[row1][col2 + 1] - f[row2 + 1][col1];
}
private final int[][] f;
}