304. Range Sum Query 2D - Immutable java solutions

Posted 2020-08-02 Miller1991

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.

该题在编程之美中出现过了。可以将矩阵看做是几何中的方形，按照求面积的方式来求子矩阵的和。主要是要处理好边界index。

可以和 一维的题目进行对比。 链接：

303. Range Sum Query - Immutable java solutions

 1 public class NumMatrix {
 2     private int[][] sum;
 3     boolean hasnum  =true;
 4     public NumMatrix(int[][] matrix) {
 5         if(matrix.length == 0 || matrix[0].length == 0){
 6             hasnum = false;
 7             return;
 8         }
 9         sum = new int[matrix.length+1][matrix[0].length+1];
10         for(int i = 0; i < sum.length; i++){
11             sum[i][0] = 0;
12         }
13         for(int i = 1; i < sum[0].length; i++){
14             sum[0][i] = 0;
15         }
16         for(int i = 1; i < sum.length; i++){
17             for(int j = 1; j < sum[0].length; j++){
18                 sum[i][j] = sum[i-1][j] + sum[i][j-1] - sum[i-1][j-1] + matrix[i-1][j-1];
19             }
20         }
21     }
22 
23     public int sumRegion(int row1, int col1, int row2, int col2) {
24         if(!hasnum) return 0;
25         return sum[row2+1][col2+1] - sum[row2+1][col1] - sum[row1][col2+1] + sum[row1][col1];
26     }
27 }
28 
29 
30 // Your NumMatrix object will be instantiated and called as such:
31 // NumMatrix numMatrix = new NumMatrix(matrix);
32 // numMatrix.sumRegion(0, 1, 2, 3);
33 // numMatrix.sumRegion(1, 2, 3, 4);