LeetCode Range Sum Query - Mutable

Posted 2020-06-18 Dylan_Java_NYC

原题链接在这里：https://leetcode.com/problems/range-sum-query-2d-mutable/

题目：

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
update(3, 2, 2)
sumRegion(2, 1, 4, 3) -> 10

题解：

用到了二维binary index tree. 

Time Complexity: builder O(mnlogmn). update O(logmn). sumRange O(logmn). m = matrix.length. n = matrix[0].length.

Space : O(mn).

AC Java:

 1 public class NumMatrix {
 2     int [][] bit;
 3     int [][] matrix;
 4     
 5     public NumMatrix(int[][] matrix) {
 6         if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
 7             return;
 8         }
 9         int m = matrix.length;
10         int n = matrix[0].length;
11         this.bit = new int[m+1][n+1];
12         this.matrix = new int[m][n];
13         for(int i = 0; i<m; i++){
14             for(int j = 0; j<n; j++){
15                 update(i, j, matrix[i][j]);
16             }
17         }
18     }
19     
20     public void update(int row, int col, int val) {
21         int diff = val - this.matrix[row][col];
22         this.matrix[row][col] = val;
23         for(int i = row+1; i<bit.length; i+=(i&-i)){
24             for(int j = col+1; j<bit[0].length; j+=(j&-j)){
25                 this.bit[i][j] += diff;
26             }
27         }
28     }
29     
30     public int sumRegion(int row1, int col1, int row2, int col2) {
31         return getSum(row2+1, col2+1) - getSum(row1, col2+1) - getSum(row2+1, col1) + getSum(row1, col1);
32     }
33     
34     private int getSum(int row, int col){
35         int sum = 0;
36         for(int i = row; i>0; i-=(i&-i)){
37             for(int j = col; j>0; j-=(j&-j)){
38                 sum += this.bit[i][j];
39             }
40         }
41         return sum;
42     }
43 }
44 
45 /**
46  * Your NumMatrix object will be instantiated and called as such:
47  * NumMatrix obj = new NumMatrix(matrix);
48  * obj.update(row,col,val);
49  * int param_2 = obj.sumRegion(row1,col1,row2,col2);
50  */

类似Range Sum Query - Mutable.