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Range Sum Query 2D - Immutable (M)
題目
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.
題意
給定一個整數矩陣，求其中一個子矩陣的數字之和，
思路
一種方法是在 0303. Range Sum Query - Immutable 的基礎上，將矩陣中每一行的sum情況快取，求指定矩陣和時，只要累加該矩陣每一行對應的sum值即可，
更好的策略是將一維快取的方法進行推廣，進行二維快取：
\(S_{ABCD} = S_{OMCN} - S_{OMBQ} - S_{OPDN} + S_{OPAQ} = S_{OC} - S_{OB} - S_{OD} + S_{OA}\)
代碼實作
Java
一維快取
class NumMatrix {
private int[][] sum;
public NumMatrix(int[][] matrix) {
if (matrix.length != 0) {
sum = new int[matrix.length][matrix[0].length + 1];
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix[0].length; j++) {
sum[i][j + 1] += sum[i][j] + matrix[i][j];
}
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
int total = 0;
for (int i = row1; i <= row2; i++) {
total += sum[i][col2 + 1] - sum[i][col1];
}
return total;
}
}
二維快取
class NumMatrix {
private int[][] sum;
public NumMatrix(int[][] matrix) {
if (matrix.length > 0) {
int m = matrix.length, n = matrix[0].length;
sum = new int[m + 1][n + 1];
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
sum[i + 1][j + 1] = sum[i + 1][j] + sum[i][j + 1] + matrix[i][j] - sum[i][j];
}
}
}
}
public int sumRegion(int row1, int col1, int row2, int col2) {
return sum[row2 + 1][col2 + 1] - sum[row2 + 1][col1] - sum[row1][col2 + 1] + sum[row1][col1];
}
}
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