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* @lc app=leetcode.cn id=304 lang=golang * * [304] 二维区域和检索 - 矩阵不可变 * * https://leetcode.cn/problems/range-sum-query-2d-immutable/description/ *

• algorithms
• Medium (59.62%)
• Likes: 419
• Dislikes: 0
• Total Accepted: 99.7K
• Total Submissions: 167K
• Testcase Example: '["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]]'

* * 给定一个二维矩阵 matrix，以下类型的多个请求： * * * 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) 。 * * * 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 。 * * * * * 示例 1： * * * * * 输入: * ["NumMatrix","sumRegion","sumRegion","sumRegion"] * * [[[[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]] * 输出: * [null, 8, 11, 12] * * 解释: * NumMatrix 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]]); * numMatrix.sumRegion(2, 1, 4, 3); // return 8 (红色矩形框的元素总和) * numMatrix.sumRegion(1, 1, 2, 2); // return 11 (绿色矩形框的元素总和) * numMatrix.sumRegion(1, 2, 2, 4); // return 12 (蓝色矩形框的元素总和) * * * * * 提示： * * * m == matrix.length * n == matrix[i].length * 1 <= m, n <= 200 * -10^5 <= matrix[i][j] <= 10^5 * 0 <= row1 <= row2 < m * 0 <= col1 <= col2 < n * 最多调用 10^4 次 sumRegion 方法 * *

Index

• type NumMatrix
  • func Constructor(matrix [][]int) NumMatrix
  • func (this *NumMatrix) SumRegion(row1 int, col1 int, row2 int, col2 int) int

Types

type NumMatrix

type NumMatrix struct {
}

@lc code=start

func Constructor

func Constructor(matrix [][]int) NumMatrix

func (*NumMatrix) SumRegion

func (this *NumMatrix) SumRegion(row1 int, col1 int, row2 int, col2 int) int