Question

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

方法一

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)

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);
 */

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);
 */

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);
 */

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)
 */