lintcode(665)Range Sum Query 2D - Immutable

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#java #lintcode

本文介绍了一种使用动态规划的方法来高效地计算二维矩阵中任意指定矩形区域内的元素总和。通过预先计算并存储部分和，可以在常数时间内响应多次查询请求。

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Description:

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).

给一个二维数组，给定4个整数作为角标(row1 , col1) (row2 , col2),返回两个角标所确定矩形内的元素之和。

注意事项

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.

Explanation:

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

Solution:

Dynamic programming.

public class NumMatrix {
    int[][] sum;
    public NumMatrix(int[][] matrix) {
        this.sum = new int[matrix.length + 1][matrix[0].length + 1];
        for(int i = 1 ; i <= matrix.length ; i++){
            for(int j = 1 ; j <= matrix[0].length ; j++){
                this.sum[i][j] = matrix[i - 1][j - 1] + this.sum[i - 1][j] + 
                                 this.sum[i][j - 1] - this.sum[i - 1][j - 1];
            }
        }
    }
    
    public int sumRegion(int row1, int col1, int row2, int col2) {
        return this.sum[row2 + 1][col2 + 1] 
        				+ this.sum[row1][col1] 
        				- this.sum[row1][col2 + 1] 
        				- this.sum[row2 + 1][col1];
    }
}

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