dp/range-sum-query-2d-immutable

Range Sum Query 2D - Immutable

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).

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

思路跟一维的类似，建立一个累加和矩阵。令状态f[i][j]表示从(0,0)到(i,j)的子矩阵的和，则状态转移方程为

f[i][j] = f[i-1][j] + rowSum

其中 rowSum 是矩阵matrix[i][0]到matrix[i][j]这一行的和。

有了f[i][j]， 则

sumRange(i1,j1,i2,j2) = f[i2][j2] + f[i1-1][j1-1] - f[i1-1][j2]-f[i2][j1-1]

将辅助矩阵f[i][j]的行数和列数增 1，可以简化对矩阵边界的处理。

代码

// Range Sum Query 2D - Immutable
public class NumMatrix {
    // Time Complexity: O(n*m), Space Complexity: O(1)
    public NumMatrix(int[][] matrix) {
        final int m = matrix.length;
        final int n = matrix.length > 0 ? matrix[0].length : 0;
        this.f = new int[m + 1][n + 1];

        for (int i = 1; i < m + 1; ++i) {
            int rowSum = 0;
            for (int j = 1; j < n + 1; ++j) {
                f[i][j] += rowSum + matrix[i-1][j-1];
                if (i > 1) {
                    f[i][j] += f[i-1][j];
                }
                rowSum += matrix[i-1][j-1];
            }
        }
    }

    // Time Complexity: O(1), Space Complexity: O(1)
    public int sumRegion(int row1, int col1, int row2, int col2) {
        return f[row2 + 1][col2 + 1] + f[row1][col1] -
                f[row1][col2 + 1] - f[row2 + 1][col1];
    }
    private final int[][] f;
}

相关题目

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