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文章目录
- 1476. 子矩形查询：
- 样例 1
- 样例 2
- 提示
- 分析
- 题解
- java
- c
- c++
- python
- go
- rust
- 原题传送门：https://leetcode-cn.com/problems/subrectangle-queries/
1476. 子矩形查询：
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：
- 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。
- getValue(int row, int col)
- 返回矩形中坐标 (row,col) 的当前值。
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样例 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
分析
- 矩形的初始值肯定要保存下来。
- 然后最直观的方式就是每次更新操作就把矩形中的值老老实实更新掉，取值就直接返回就可以了。
- 根据提示，我们可以知道矩形最大是100 * 100，也就是每次更新值最多可能需要更新10000个值，但是更新次数最多会有500次，所以我们可以记录每次更新操作，在取值的时候我们倒着查找历史操作，如果点没有做过更新，那就是原值。这样更新操作的时间复杂度仅仅是O(1)，而取值操作最多就是500次循环去查找历史。
- 到底哪种方式好不是绝对的，可以根据实际情况，本题的情况是值多，操作少，所以我们不做更新，而是把历史操作存下来。
题解 java
class SubrectangleQueries {
private final int[][] rectangle;
private List history = new ArrayList<>();
public SubrectangleQueries(int[][] rectangle) {
this.rectangle = rectangle;
}
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[] h = history.get(i);
if (row >= h[0] && row <= h[2]
&& col >= h[1] && col <= h[3]) {
return h[4];
}
}
return rectangle[row][col];
}
}
c
typedef struct {
int value[100][100];
int history[500][5];
int history_size;
} SubrectangleQueries;
SubrectangleQueries *subrectangleQueriesCreate(int **rectangle, int rectangleSize, int *rectangleColSize) {
SubrectangleQueries *obj = NULL;
obj = (SubrectangleQueries *) malloc(sizeof(SubrectangleQueries));
if (obj == NULL) {
return NULL;
}
for (int i = 0; i < rectangleSize; ++i) {
for (int j = 0; j < rectangleColSize[i]; ++j) {
obj->value[i][j] = rectangle[i][j];
}
}
return obj;
}
void
subrectangleQueriesUpdateSubrectangle(SubrectangleQueries *obj, int row1, int col1, int row2, int col2, int newValue) {
int *h = obj->history[obj->history_size++];
h[0] = row1;
h[1] = col1;
h[2] = row2;
h[3] = col2;
h[4] = newValue;
}
int subrectangleQueriesGetValue(SubrectangleQueries *obj, int row, int col) {
for (int i = obj->history_size - 1; i >= 0; --i) {
int *h = obj->history[i];
if (row >= h[0] && row <= h[2]
&& col >= h[1] && col <= h[3]) {
return h[4];
}
}
return obj->value[row][col];
}
void subrectangleQueriesFree(SubrectangleQueries *obj) {
free(obj);
}
c++
class SubrectangleQueries {
private:
vector> rectangle;
vector> history;
public:
SubrectangleQueries(vector>& rectangle) : rectangle(rectangle) {
}
void updateSubrectangle(int row1, int col1, int row2, int col2, int newValue) {
history.push_back({row1, col1, row2, col2, newValue});
}
int getValue(int row, int col) {
for (int i = history.size() - 1; i >= 0; --i) {
vector h = history[i];
if (row >= h[0] && row <= h[2]
&& col >= h[1] && col <= h[3]) {
return h[4];
}
}
return rectangle[row][col];
}
};
python
class SubrectangleQueries:
def __init__(self, rectangle: List[List[int]]):
self.rectangle = 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 i in range(len(self.history) - 1, -1, -1):
row1, col1, row2, col2, val = self.history[i]
if row1 <= row <= row2 and col1 <= col <= col2:
return val
return self.rectangle[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)
go
type SubrectangleQueries struct {
rectangle [][]int
history [][]int
}
func Constructor(rectangle [][]int) SubrectangleQueries {
return SubrectangleQueries{rectangle: rectangle}
}
func (this *SubrectangleQueries) UpdateSubrectangle(row1 int, col1 int, row2 int, col2 int, newValue int) {
this.history = append(this.history, []int{row1, col1, row2, col2, newValue})
}
func (this *SubrectangleQueries) GetValue(row int, col int) int {
for i := len(this.history) - 1; i >= 0; i-- {
h := this.history[i]
if h[0] <= row && row <= h[2] && h[1] <= col && col <= h[3] {
return h[4]
}
}
return this.rectangle[row][col]
}
rust
struct SubrectangleQueries {
rectangle: Vec>,
history: Vec<(i32, i32, i32, i32, i32)>,
}
impl SubrectangleQueries {
fn new(rectangle: Vec>) -> Self {
Self {rectangle, history: vec![]}
}
fn update_subrectangle(&mut self, row1: i32, col1: i32, row2: i32, col2: i32, new_value: i32) {
self.history.push((row1, col1, row2, col2, new_value));
}
fn get_value(&self, row: i32, col: i32) -> i32 {
for &(r1, c1, r2, c2, val) in self.history.iter().rev() {
if r1 <= row && row <= r2 && c1 <= col && col <= c2 {
return val
}
}
self.rectangle[row as usize][col as usize]
}
}
原题传送门：https://leetcode-cn.com/problems/subrectangle-queries/
