题目描述

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.

样例

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

算法

class SubrectangleQueries {
    vector<vector<int>> matrix;
public:
    SubrectangleQueries(vector<vector<int>>& rectangle) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        int m = rectangle.size(), n = rectangle[0].size();
        matrix.resize(m, vector<int>(n));
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                matrix[i][j] = rectangle[i][j];
            }
        }
    }

    void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
        for (int i = row1; i <= row2; ++i) {
            for (int j = col1; j <= col2; ++j) {
                matrix[i][j] = newValue;
            }
        }
    }

    int getValue(int row, int col) {
        return matrix[row][col];
    }
};

/**
 * Your SubrectangleQueries object will be instantiated and called as such:
 * SubrectangleQueries* obj = new SubrectangleQueries(rectangle);
 * obj->updateSubrectangle(row1,col1,row2,col2,newValue);
 * int param_2 = obj->getValue(row,col);
 */

updateSubrectangle操作是把给定范围内矩形内的元素全都更新成newvalue，而不是加上newValue，如果是加上newValue，那么就可以使用延迟修改了。这里我们就直接遍历修改好了。

最坏的情况，每次都需要修改矩阵的所有元素，$500 \times 100 \times 100$，不会超时，每次查询getValue都是$O(1)$。