Range Sum Query 2D - Immutable 计算矩形中的元素和
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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) -> 8sumRegion(1, 1, 2, 2) -> 11sumRegion(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.
我们先来考虑一下如何计算(2, 1, 4, 3)。
我们设完整的矩形total为:以(0,0)为左上角点(4,3)为右下角点组成的矩形。
左边的矩形left:以(0,0) (4,1)组成的矩形。
上面的矩形top:以(0, 0) (2, 3) 组成的矩形。
左上角的矩形leftTop: 以(0, 0)(2,1)组成的矩形。
目标矩形的sum可以表示为 sum_of_total - sum_of_left - sum_of_right + sum_of_leftTop(因为它被减了2次)。
如果我们有一个矩形,它的i, j 下标存储了以(0, 0 ) 为左上角点,以( i , j ) 为右下角点的矩形的和。
那么计算每一个查询的sum 的时间复杂度为O(1)。
运行时间:
为了避免边界check问题。
我们开的cache矩形为[m + 1] [ n + 1]大小的。它的边界值就是0。
代码:
public class RangeSumQuery2DImmutable { private int[][] cache = new int[0][0]; public RangeSumQuery2DImmutable(int[][] matrix) { int m = matrix.length; if (m == 0) { return; } int n = matrix[0].length; cache = new int[m + 1][n + 1]; for (int i = 1; i <= m; i++) { for (int j = 1; j <=n; j++) { cache[i][j] = cache[i][j - 1] + cache[i - 1][j] - cache[i - 1][j - 1] + matrix[i - 1][j - 1]; } } } public int sumRegion(int row1, int col1, int row2, int col2) { row2++; col2++; int total = cache[row2][col2]; int left = cache[row2][col1]; int top = cache[row1][col2]; int leftTop = cache[row1][col1]; return total - left - top + leftTop; }}
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