0304. Range Sum Query 2 D Immutable

304. Range Sum Query 2D - Immutable #

题目 #

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).

https://leetcode.com/static/images/courses/range_sum_query_2d.png

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:

  1. You may assume that the matrix does not change.
  2. There are many calls to sumRegion function.
  3. You may assume that row1 ≤ row2 and col1 ≤ col2.

题目大意 #

给定一个二维矩阵,计算其子矩形范围内元素的总和,该子矩阵的左上角为 (row1, col1) ,右下角为 (row2, col2) 。

解题思路 #

  • 这一题是一维数组前缀和的进阶版本。定义 f(x,y) 代表矩形左上角 (0,0),右下角 (x,y) 内的元素和。 \[ f(i,j) = \sum_{x=0}^{i}\sum_{y=0}^{j} Matrix[x][y]\]

    \[ \begin{aligned} f(i,j) &= \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{x=0}^{i-1} Matrix[x][j] + \sum_{y=0}^{j-1} Matrix[i][y] + Matrix[i][j]\\ &= (\sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{x=0}^{i-1} Matrix[x][j]) + (\sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + \sum_{y=0}^{j-1} Matrix[i][y]) - \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + Matrix[i][j]\\ &= \sum_{x=0}^{i-1}\sum_{y=0}^{j} Matrix[x][y] + \sum_{x=0}^{i}\sum_{y=0}^{j-1} Matrix[x][y] - \sum_{x=0}^{i-1}\sum_{y=0}^{j-1} Matrix[x][y] + Matrix[i][j]\\ &= f(i-1,j) + f(i,j-1) - f(i-1,j-1) + Matrix[i][j] \end{aligned} \]
  • 于是得到递推的关系式:f(i, j) = f(i-1, j) + f(i, j-1) - f(i-1, j-1) + matrix[i][j],写代码为了方便,新建一个 m+1 * n+1 的矩阵,这样就不需要对 row = 0col = 0 做单独处理了。上述推导公式如果画成图也很好理解:

    https://img.halfrost.com/Leetcode/leetcode_304.png

    左图中大的矩形由粉红色的矩形 + 绿色矩形 - 粉红色和绿色重叠部分 + 黄色部分。这就对应的是上面推导出来的递推公式。左图是矩形左上角为 (0,0) 的情况,更加一般的情况是右图,左上角是任意的坐标,公式不变。

  • 时间复杂度:初始化 O(mn),查询 O(1)。空间复杂度 O(mn)

代码 #

package leetcode

type NumMatrix struct {
	cumsum [][]int
}

func Constructor(matrix [][]int) NumMatrix {
	if len(matrix) == 0 {
		return NumMatrix{nil}
	}
	cumsum := make([][]int, len(matrix)+1)
	cumsum[0] = make([]int, len(matrix[0])+1)
	for i := range matrix {
		cumsum[i+1] = make([]int, len(matrix[i])+1)
		for j := range matrix[i] {
			cumsum[i+1][j+1] = matrix[i][j] + cumsum[i][j+1] + cumsum[i+1][j] - cumsum[i][j]
		}
	}
	return NumMatrix{cumsum}
}

func (this *NumMatrix) SumRegion(row1 int, col1 int, row2 int, col2 int) int {
	cumsum := this.cumsum
	return cumsum[row2+1][col2+1] - cumsum[row1][col2+1] - cumsum[row2+1][col1] + cumsum[row1][col1]
}

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

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