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子矩形查询
请你实现一个类 SubrectangleQueries ，它的构造函数的参数是一个 rows x cols 的矩形（这里用整数矩阵表示），并支持以下两种操作： 1. updateSubrectangle(int row1, int col1, int row2, int col2, int newV…
3
题型
5
代码语言
3
相关题
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中等 · 数组·结合·design
答案摘要
class SubrectangleQueries: def __init__(self, rectangle: List[List[int]]):
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题目描述
请你实现一个类 SubrectangleQueries ，它的构造函数的参数是一个 rows x cols 的矩形（这里用整数矩阵表示），并支持以下两种操作：
1. updateSubrectangle(int row1, int col1, int row2, int col2, int newValue)
- 用
newValue更新以(row1,col1)为左上角且以(row2,col2)为右下角的子矩形。
2. getValue(int row, int col)
- 返回矩形中坐标
(row,col)的当前值。
示例 1：
输入： ["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","updateSubrectangle","getValue","getValue"] [[[[1,2,1],[4,3,4],[3,2,1],[1,1,1]]],[0,2],[0,0,3,2,5],[0,2],[3,1],[3,0,3,2,10],[3,1],[0,2]] 输出： [null,1,null,5,5,null,10,5] 解释： SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,2,1],[4,3,4],[3,2,1],[1,1,1]]); // 初始的 (4x3) 矩形如下： // 1 2 1 // 4 3 4 // 3 2 1 // 1 1 1 subrectangleQueries.getValue(0, 2); // 返回 1 subrectangleQueries.updateSubrectangle(0, 0, 3, 2, 5); // 此次更新后矩形变为： // 5 5 5 // 5 5 5 // 5 5 5 // 5 5 5 subrectangleQueries.getValue(0, 2); // 返回 5 subrectangleQueries.getValue(3, 1); // 返回 5 subrectangleQueries.updateSubrectangle(3, 0, 3, 2, 10); // 此次更新后矩形变为： // 5 5 5 // 5 5 5 // 5 5 5 // 10 10 10 subrectangleQueries.getValue(3, 1); // 返回 10 subrectangleQueries.getValue(0, 2); // 返回 5
示例 2：
输入： ["SubrectangleQueries","getValue","updateSubrectangle","getValue","getValue","updateSubrectangle","getValue"] [[[[1,1,1],[2,2,2],[3,3,3]]],[0,0],[0,0,2,2,100],[0,0],[2,2],[1,1,2,2,20],[2,2]] 输出： [null,1,null,100,100,null,20] 解释： SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,1,1],[2,2,2],[3,3,3]]); subrectangleQueries.getValue(0, 0); // 返回 1 subrectangleQueries.updateSubrectangle(0, 0, 2, 2, 100); subrectangleQueries.getValue(0, 0); // 返回 100 subrectangleQueries.getValue(2, 2); // 返回 100 subrectangleQueries.updateSubrectangle(1, 1, 2, 2, 20); subrectangleQueries.getValue(2, 2); // 返回 20
提示：
- 最多有
500次updateSubrectangle和getValue操作。 1 <= rows, cols <= 100rows == rectangle.lengthcols == rectangle[i].length0 <= row1 <= row2 < rows0 <= col1 <= col2 < cols1 <= newValue, rectangle[i][j] <= 10^90 <= row < rows0 <= col < cols
解题思路
方法一
class SubrectangleQueries:
def __init__(self, rectangle: List[List[int]]):
self.g = rectangle
self.ops = []
def updateSubrectangle(
self, row1: int, col1: int, row2: int, col2: int, newValue: int
) -> None:
self.ops.append((row1, col1, row2, col2, newValue))
def getValue(self, row: int, col: int) -> int:
for r1, c1, r2, c2, v in self.ops[::-1]:
if r1 <= row <= r2 and c1 <= col <= c2:
return v
return self.g[row][col]
# Your SubrectangleQueries object will be instantiated and called as such:
# obj = SubrectangleQueries(rectangle)
# obj.updateSubrectangle(row1,col1,row2,col2,newValue)
# param_2 = obj.getValue(row,col)
复杂度分析
| 指标 | 值 |
|---|---|
| 时间 | Depends on the final approach |
| 空间 | Depends on the final approach |
面试官常问的追问
外企场景- question_mark
Focus on how well the candidate handles large updates in a matrix.
- question_mark
Look for any optimizations made to avoid redundant operations.
- question_mark
Test the candidate's ability to balance time and space complexity when designing the solution.
常见陷阱
外企场景- error
Failing to optimize the updateSubrectangle method for larger input sizes, which may result in time inefficiencies.
- error
Overcomplicating the solution with unnecessary space optimizations without considering the problem's constraints.
- error
Assuming that every update is independent when they may overlap, leading to wasted operations or errors.
进阶变体
外企场景- arrow_right_alt
Allow for dynamic resizing of the rectangle after multiple updates.
- arrow_right_alt
Support for handling multiple different types of updates simultaneously.
- arrow_right_alt
Extend the functionality to support queries on entire rows or columns.