308.Range-Sum-Query-2D---Mutable

308. Range Sum Query 2D Mutable

题目地址

https://leetcode.com/problems/range-sum-query-2d-mutable/

题目描述

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).

Range Sum Query 2D
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
Note:
The matrix is only modifiable by the update function.
You may assume the number of calls to update and sumRegion function is distributed evenly.
You may assume that row1 ≤ row2 and col1 ≤ col2.

代码

Approach #1 Binary Indexed Tree (BIT)

https://www.topcoder.com/community/competitive-programming/tutorials/binary-indexed-trees/

一个特殊的位运算： 获取数字 A 的最低有效位：A & -A 或者 A & ~(A - 1) 例如数字 6（110）的最低位对应 2（10）

清除数字 A 的最低有效位：A = A - (A & -A) 或者 A = A - (A & ~(A - 1)) 例如数字 6（110）的最低位对应 2（10），清除后得到 4（100）

tree[i][j] saves the rangeSum of [i-(i&-i), i] x [j-(j&-j), j]

class NumMatrix {
    int[][] tree;
  int[][] nums;
  int m;
  int n;

  public NumMatrix(int[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0)    return;
    m = matrix.length;
    n = matrix[0].length;
    tree = new int[m + 1][n + 1];
    nums = new int[m][n];
    for (int i = 0; i < m; i++) {
      for (int j = 0; j < n; j++) {
        update(i, j, matrix[i][j]);
      }
    }
  }

  public void update(int row, int col, int val) {
        if (m == 0 || n == 0)        return;
    int delta = val - nums[row][col];
    nums[row][col] = val;
    for (int i = row + 1; i <= m; i += i & (-i)) {
      for (int j = col + 1; j<= n; j += j & (-j)) {
        tree[i][j] += delta;
      }
    }
  }

  public int sumRegion(int row1, int col1, int row2, int col2) {
        if (m == 0 || n == 0)    return 0;
    return sum(row2 + 1, col2 + 1) + sum(row1, col1) - sum (row1, col2 + 1) - sum(row2 + 1, col1);
  }

  private int sum(int row, int col) {
    int sum = 0;
    for (int i = row; i > 0; i -= i & (-i)) {
      for (int j = col; j > 0; j -= j & (-j)) {
        sum += tree[i][j];
      }
    }
    return sum;
  }

}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * obj.update(row,col,val);
 * int param_2 = obj.sumRegion(row1,col1,row2,col2);
 */