刷题-力扣-1476. 子矩形查询

1476. 子矩形查询

题目链接

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

题目分析

1. 根据题目描述，私有成员变量matrix存储矩阵
2. 成员函数updateSubrectangle内使用双重循环修改矩阵的值
3. 成员函数getValue返回矩阵指定位置的值

代码

class SubrectangleQueries {
private:
    vector<vector<int>> matrix;
public:
    SubrectangleQueries(vector<vector<int>>& rectangle) {
        matrix = rectangle;
    }
    
    void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
        for (int row = row1; row <= row2; row++) {
            for (int col = col1; col <= col2; col++) {
                matrix[row][col] = 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);
 */