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1476. 子矩形查询

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

方法一

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class SubrectangleQueries:
    def __init__(self, rectangle: List[List[int]]):
        self.g = rectangle
        self.ops = []

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

    def getValue(self, row: int, col: int) -> int:
        for r1, c1, r2, c2, v in self.ops[::-1]:
            if r1 <= row <= r2 and c1 <= col <= c2:
                return v
        return self.g[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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class SubrectangleQueries {
    private int[][] g;
    private LinkedList<int[]> ops = new LinkedList<>();

    public SubrectangleQueries(int[][] rectangle) {
        g = rectangle;
    }

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

    public int getValue(int row, int col) {
        for (var op : ops) {
            if (op[0] <= row && row <= op[2] && op[1] <= col && col <= op[3]) {
                return op[4];
            }
        }
        return g[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);
 */
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class SubrectangleQueries {
public:
    vector<vector<int>> g;
    vector<vector<int>> ops;

    SubrectangleQueries(vector<vector<int>>& rectangle) {
        g = rectangle;
    }

    void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
        ops.push_back({row1, col1, row2, col2, newValue});
    }

    int getValue(int row, int col) {
        for (int i = ops.size() - 1; ~i; --i) {
            auto op = ops[i];
            if (op[0] <= row && row <= op[2] && op[1] <= col && col <= op[3]) {
                return op[4];
            }
        }
        return g[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);
 */
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type SubrectangleQueries struct {
    g   [][]int
    ops [][]int
}

func Constructor(rectangle [][]int) SubrectangleQueries {
    return SubrectangleQueries{rectangle, [][]int{}}
}

func (this *SubrectangleQueries) UpdateSubrectangle(row1 int, col1 int, row2 int, col2 int, newValue int) {
    this.ops = append(this.ops, []int{row1, col1, row2, col2, newValue})
}

func (this *SubrectangleQueries) GetValue(row int, col int) int {
    for i := len(this.ops) - 1; i >= 0; i-- {
        op := this.ops[i]
        if op[0] <= row && row <= op[2] && op[1] <= col && col <= op[3] {
            return op[4]
        }
    }
    return this.g[row][col]
}

/**
 * Your SubrectangleQueries object will be instantiated and called as such:
 * obj := Constructor(rectangle);
 * obj.UpdateSubrectangle(row1,col1,row2,col2,newValue);
 * param_2 := obj.GetValue(row,col);
 */
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class SubrectangleQueries {
    g: number[][];
    ops: number[][];
    constructor(rectangle: number[][]) {
        this.g = rectangle;
        this.ops = [];
    }

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

    getValue(row: number, col: number): number {
        for (let i = this.ops.length - 1; ~i; --i) {
            const [r1, c1, r2, c2, v] = this.ops[i];
            if (r1 <= row && row <= r2 && c1 <= col && col <= c2) {
                return v;
            }
        }
        return this.g[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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