304.Range-Sum-Query-2D---Immutable

304. Range Sum Query 2D Immutable

题目地址

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

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
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.

代码

Approach #1 Caching Rows

Time: O(mn) && Space: O(mn)

class NumMatrix {
    private int[][] dp;
  public NumMatrix(int[][] matrix) {
        if (matrix.length == 0 || matrix[0].length == 0)        return;
    dp = new int[matrix.length][matrix[0].length + 1];
    for (int r = 0; r < matrix.length; r++) {
      for (int c = 0; c < matrix[0].length; c++) {
        dp[r][c+1] = dp[r][c] + matrix[r][c];
      }
    }
  }

  public int sumRegion(int row1, int col1, int row2, int col2) {
        int sum = 0;
    for (int row = row1; row <= row2; row++) {
      sum += dp[row][col2+1] - dp[row][col1];
    }
    return sum;
  }
}

/**
 * Your NumMatrix object will be instantiated and called as such:
 * NumMatrix obj = new NumMatrix(matrix);
 * int param_1 = obj.sumRegion(row1,col1,row2,col2);
 */

Approach #2 Caching Smarter

• Time complexity : O(1) time per query, O(mn) time pre-computation. The pre-computation in the constructor takes O(mn) time. Each sumRegion query takes O(1) time.

• Space complexity : O(mn). The algorithm uses O(mn) space to store the cumulative region sum.

class Solution {
  private int[][] dp;
  public NumMatrix(int[][] matrix) {
    if (matrix.length == 0 || matrix[0].length == 0)        return;
    dp = new int[matrix.length + 1][matrix[0].length + 1];
    for (int r = 0; r < matrix.length; r++) {
      for (int c = 0; c < matrix[0].length; c++) {
        dp[r+1][c+1] = dp[r+1][c] + dp[r][c+1] + matrix[r][c] - dp[r][c];
      }
    }
  }

  public int sumRegion(int row1, int col1, int row2, int col2) {
    return dp[row2+1][col2+1] - dp[row1][col2+1] - dp[row2+1][col1] + dp[row1][col1];
  }

}