304. Range Sum Query 2D - Immutable

Leetcode Dynamic Programming

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.

分析

这道题目是LeetCode 303. Range Sum Query - Immutable的加强版，关键是找到动态规划的状态转移方程。

一种非常巧妙的方法是将矩形的和，转换为一系列矩形的和，用sum()表示矩形的面积，例如图

中的矩形ABCD可以用OD、OB、OC、OA来表示：

Sum(ABCD)=Sum(OD)−Sum(OB)−Sum(OC)+Sum(OA)

这样一来，只要将以(0,0)为左上角，(i,j)为右下角的矩形的面积保存起来，任意两点构造的矩形的面积就非常容易求解了。

class NumMatrix {
    private int[][] regionSum;
    public NumMatrix(int[][] matrix) {
        int m = matrix.length, n = m > 0 ? matrix[0].length : 0;
        regionSum = new int[m + 1][n + 1];
        for (int i = 1; i <= m; i++)
            for (int j = 1; j <= n; j++)
                regionSum[i][j] = regionSum[i - 1][j] + regionSum[i][j - 1] - 
                regionSum[i - 1][j - 1] + matrix[i - 1][j - 1];
    }

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