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  • 【leetcode】1476. Subrectangle Queries

    题目如下:

    Implement the class SubrectangleQueries which receives a rows x cols rectangle as a matrix of integers in the constructor and supports two methods:

    1. updateSubrectangle(int row1, int col1, int row2, int col2, int newValue)

    • Updates all values with newValue in the subrectangle whose upper left coordinate is (row1,col1) and bottom right coordinate is (row2,col2).

    2. getValue(int row, int col)

    • Returns the current value of the coordinate (row,col) from the rectangle. 

    Example 1:

    Input
    ["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]]
    Output
    [null,1,null,5,5,null,10,5]
    Explanation
    SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,2,1],[4,3,4],[3,2,1],[1,1,1]]);  
    // The initial rectangle (4x3) looks like:
    // 1 2 1
    // 4 3 4
    // 3 2 1
    // 1 1 1
    subrectangleQueries.getValue(0, 2); // return 1
    subrectangleQueries.updateSubrectangle(0, 0, 3, 2, 5);
    // After this update the rectangle looks like:
    // 5 5 5
    // 5 5 5
    // 5 5 5
    // 5 5 5 
    subrectangleQueries.getValue(0, 2); // return 5
    subrectangleQueries.getValue(3, 1); // return 5
    subrectangleQueries.updateSubrectangle(3, 0, 3, 2, 10);
    // After this update the rectangle looks like:
    // 5   5   5
    // 5   5   5
    // 5   5   5
    // 10  10  10 
    subrectangleQueries.getValue(3, 1); // return 10
    subrectangleQueries.getValue(0, 2); // return 5
    

    Example 2:

    Input
    ["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]]
    Output
    [null,1,null,100,100,null,20]
    Explanation
    SubrectangleQueries subrectangleQueries = new SubrectangleQueries([[1,1,1],[2,2,2],[3,3,3]]);
    subrectangleQueries.getValue(0, 0); // return 1
    subrectangleQueries.updateSubrectangle(0, 0, 2, 2, 100);
    subrectangleQueries.getValue(0, 0); // return 100
    subrectangleQueries.getValue(2, 2); // return 100
    subrectangleQueries.updateSubrectangle(1, 1, 2, 2, 20);
    subrectangleQueries.getValue(2, 2); // return 20

    Constraints:

    • There will be at most 500 operations considering both methods: updateSubrectangle and getValue.
    • 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

    解题思路:本题的数据量不大,O(n^2)的复杂度完全可行。

    代码如下:

    class SubrectangleQueries(object):
    
        def __init__(self, rectangle):
            """
            :type rectangle: List[List[int]]
            """
            self.rec = rectangle
    
        def updateSubrectangle(self, row1, col1, row2, col2, newValue):
            """
            :type row1: int
            :type col1: int
            :type row2: int
            :type col2: int
            :type newValue: int
            :rtype: None
            """
            for i in range(row1,row2+1):
                for j in range(col1,col2+1):
                    self.rec[i][j] = newValue
    
    
        def getValue(self, row, col):
            """
            :type row: int
            :type col: int
            :rtype: int
            """
            return self.rec[row][col]
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  • 原文地址:https://www.cnblogs.com/seyjs/p/13585552.html
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