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1476. 子矩形查询

English Version

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

解法

用历史记录列表保存修改历史。

Python3

class SubrectangleQueries:

    def __init__(self, rectangle: List[List[int]]):
        self.rec = rectangle
        self.history = []

    def updateSubrectangle(self, row1: int, col1: int, row2: int, col2: int, newValue: int) -> None:
        self.history.append((row1, col1, row2, col2, newValue))

    def getValue(self, row: int, col: int) -> int:
        for row1, col1, row2, col2, newValue in self.history[::-1]:
            if row >= row1 and row <= row2 and col >= col1 and col <= col2:
                return newValue
        return self.rec[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)

Java

class SubrectangleQueries {
    private int[][] rec;
    private List<int[]> history;

    public SubrectangleQueries(int[][] rectangle) {
        rec = rectangle;
        history = new ArrayList<>();
    }

    public void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
        history.add(new int[]{row1, col1, row2, col2, newValue});
    }

    public int getValue(int row, int col) {
        for (int i = history.size() - 1; i >= 0; --i) {
            int[] record = history.get(i);
            if (row >= record[0] && row <= record[2] && col >= record[1] && col <= record[3]) {
                return record[4];
            }
        }
        return rec[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);
 */

TypeScript

class SubrectangleQueries {
    grid: number[][];
    history: number[][];
    constructor(rectangle: number[][]) {
        this.grid = rectangle;
        this.history = [];
    }

    updateSubrectangle(
        row1: number,
        col1: number,
        row2: number,
        col2: number,
        newValue: number
    ): void {
        this.history.push([row1, col1, row2, col2, newValue]);
    }

    getValue(row: number, col: number): number {
        for (let i = this.history.length - 1; i >= 0; --i) {
            let [row1, col1, row2, col2, newValue] = this.history[i];
            if (row >= row1 && row <= row2 && col >= col1 && col <= col2) {
                return newValue;
            }
        }
        return this.grid[row][col];
    }
}

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

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