{ "data": { "question": { "questionId": "304", "questionFrontendId": "304", "categoryTitle": "Algorithms", "boundTopicId": 1459, "title": "Range Sum Query 2D - Immutable", "titleSlug": "range-sum-query-2d-immutable", "content": "
Given a 2D matrix matrix
, handle multiple queries of the following type:
matrix
inside the rectangle defined by its upper left corner (row1, col1)
and lower right corner (row2, col2)
.Implement the NumMatrix
class:
NumMatrix(int[][] matrix)
Initializes the object with the integer matrix matrix
.int sumRegion(int row1, int col1, int row2, int col2)
Returns the sum of the elements of matrix
inside the rectangle defined by its upper left corner (row1, col1)
and lower right corner (row2, col2)
.You must design an algorithm where sumRegion
works on O(1)
time complexity.
\n
Example 1:
\n\n\nInput\n["NumMatrix", "sumRegion", "sumRegion", "sumRegion"]\n[[[[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]]], [2, 1, 4, 3], [1, 1, 2, 2], [1, 2, 2, 4]]\nOutput\n[null, 8, 11, 12]\n\nExplanation\nNumMatrix numMatrix = new NumMatrix([[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]]);\nnumMatrix.sumRegion(2, 1, 4, 3); // return 8 (i.e sum of the red rectangle)\nnumMatrix.sumRegion(1, 1, 2, 2); // return 11 (i.e sum of the green rectangle)\nnumMatrix.sumRegion(1, 2, 2, 4); // return 12 (i.e sum of the blue rectangle)\n\n\n
\n
Constraints:
\n\nm == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-104 <= matrix[i][j] <= 104
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
104
calls will be made to sumRegion
.给定一个二维矩阵 matrix
,以下类型的多个请求:
(row1, col1)
,右下角 为 (row2, col2)
。实现 NumMatrix
类:
NumMatrix(int[][] matrix)
给定整数矩阵 matrix
进行初始化int sumRegion(int row1, int col1, int row2, int col2)
返回 左上角 (row1, col1)
、右下角 (row2, col2)
所描述的子矩阵的元素 总和 。\n\n
示例 1:
\n\n\n\n\n输入: \n[\"NumMatrix\",\"sumRegion\",\"sumRegion\",\"sumRegion\"]\n[[[[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]]],[2,1,4,3],[1,1,2,2],[1,2,2,4]]\n输出: \n[null, 8, 11, 12]\n\n解释:\nNumMatrix numMatrix = new NumMatrix([[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]]);\nnumMatrix.sumRegion(2, 1, 4, 3); // return 8 (红色矩形框的元素总和)\nnumMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和)\nnumMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和)\n\n\n
\n\n
提示:
\n\nm == matrix.length
n == matrix[i].length
1 <= m, n <= 200
-105 <= matrix[i][j] <= 105
0 <= row1 <= row2 < m
0 <= col1 <= col2 < n
104
次 sumRegion
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