[leetcode] 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). Range Sum Query 2D 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：思路同Range Sum Query - Immutable, 只不过现在利用额外的二维数组做缓存，使后面的每次查询的时间复杂度都为O(1).
例如，原数组为：
3 0 1 5 6 3 1 2 0
构造后的缓存数组为：
0 0 0 0 0 3 3 4 0 8 14 18 0 9 17 21
比如要查询下面的区域之和：
3 0 1 5 6 3 1 2 0
则对应缓存数组中的位置：
0 0 0 0 0 3 3 4 0 8 14 18 0 9 17 21
即：21-9-4+3 = 11.
Java代码如下：
public class NumMatrix { int[][] sum; public NumMatrix(int[][] matrix) { if (matrix == null || matrix.length == 0) { return; } int m = matrix.length; int n = matrix[0].length; sum = new int[m+1][n+1]; for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { sum[i+1][j+1] = sum[i+1][j] + sum[i][j+1] - sum[i][j] + matrix[i][j]; } } } public int sumRegion(int row1, int col1, int row2, int col2) { return sum[row2+1][col2+1] - sum[row2+1][col1] - sum[row1][col2+1] + sum[row1][col1]; } }优质内容筛选与推荐>>
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