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  • 1476. 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 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 <= 100
    rows == rectangle.length
    cols == rectangle[i].length
    0 <= row1 <= row2 < rows
    0 <= col1 <= col2 < cols
    1 <= newValue, rectangle[i][j] <= 10^9
    0 <= row < rows
    0 <= col < cols

    来源:力扣(LeetCode)
    链接:https://leetcode-cn.com/problems/subrectangle-queries

    class SubrectangleQueries:
    
        def __init__(self, rectangle: List[List[int]]):
            self.data=rectangle
            self.update=[]
            
        def updateSubrectangle(self, row1: int, col1: int, row2: int, col2: int, newValue: int) -> None:
            self.update.append((row1,col1,row2,col2,newValue))
    
        def getValue(self, row: int, col: int) -> int:
            res=None
            for i in range(len(self.update)-1,-1,-1):
                row1,col1,row2,col2,value=self.update[i]
                if row1<=row<=row2 and col1<=col<=col2:
                    res=value
                    break
            return res if res else self.data[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)
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  • 原文地址:https://www.cnblogs.com/xxxsans/p/13710181.html
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