304 Range Sum Query 2D - Immutable 二维区域和检索 - 矩阵不可变
Description:
Given a 2D matrix matrix, find the sum of the elements inside the rectangle defined by its upper left corner (row1, col1) and lower right corner (row2, col2).
Example:
Given matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
]
sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12
Note:
You may assume that the matrix does not change.
There are many calls to sumRegion function.
You may assume that row1 ≤ row2 and col1 ≤ col2.
题目描述:
给定一个二维矩阵，计算其子矩形范围内元素的总和，该子矩阵的左上角为 (row1, col1) ，右下角为 (row2, col2)。
示例 :
给定 matrix = [
[3, 0, 1, 4, 2],
[5, 6, 3, 2, 1],
[1, 2, 0, 1, 5],
[4, 1, 0, 1, 7],
[1, 0, 3, 0, 5]
]
sumRegion(2, 1, 4, 3) -> 8
sumRegion(1, 1, 2, 2) -> 11
sumRegion(1, 2, 2, 4) -> 12
说明:
你可以假设矩阵不可变。
会多次调用 sumRegion 方法。
你可以假设 row1 ≤ row2 且 col1 ≤ col2。
思路:
动态规划
参考LeetCode #303 Range Sum Query - Immutable 区域和检索 - 数组不可变
在二维数组, dp[i + 1][j + 1]的前缀和表示为matrix[i][j]左上角元素之和
如
[3, 0, 1] [3, 3, 4]
[5, 6, 3] -> [8, 14, 18]
[1, 2, 0] [9, 17, 21]
dp[i + 1][j + 1] = dp[i][j + 1] + dp[i + 1][j] + matrix[i][j] - dp[i][j]
sumRegion(row1, col1, row2, col2) = dp[row2 + 1][col2 + 1] - dp[row2 + 1][col1] - dp[row1][col2 + 1] + dp[row1][col1]
这个和概率论里的联合分布有点像
时间复杂度O(1), 每次调用 sumRegion(row1, col1, row2, col2), 预处理时间复杂度O(mn), 空间复杂度O(mn)
代码:
C++:
class NumMatrix
{
private:
vector> dp;
public:
NumMatrix(vector>& matrix)
{
if (matrix.empty() or matrix[0].empty()) return;
dp.resize(matrix.size() + 1);
for (int i = 0; i < matrix.size() + 1; i++) dp[i].resize(matrix[0].size() + 1, 0);
for (int i = 0; i < matrix.size(); i++) for (int j = 0; j < matrix[0].size(); j++) dp[i + 1][j + 1] = dp[i][j + 1] + dp[i + 1][j] + matrix[i][j] - dp[i][j];
}
int sumRegion(int row1, int col1, int row2, int col2)
{
return dp[row2 + 1][col2 + 1] - dp[row1][col2 + 1] - dp[row2 + 1][col1] + dp[row1][col1];
}
};
/**
* Your NumMatrix object will be instantiated and called as such:
* NumMatrix* obj = new NumMatrix(matrix);
* int param_1 = obj->sumRegion(row1,col1,row2,col2);
*/
Java:
class NumMatrix {
private int[][] dp;
public NumMatrix(int[][] matrix) {
if (matrix.length == 0 || matrix[0].length == 0) return;
dp = new int[matrix.length + 1][matrix[0].length + 1];
for (int r = 0; r < matrix.length; r++) for (int c = 0; c < matrix[0].length; c++) dp[r + 1][c + 1] = dp[r + 1][c] + dp[r][c + 1] + matrix[r][c] - dp[r][c];
}
public int sumRegion(int row1, int col1, int row2, int col2) {
return dp[row2 + 1][col2 + 1] - dp[row1][col2 + 1] - dp[row2 + 1][col1] + dp[row1][col1];
}
}
/**
* Your NumMatrix object will be instantiated and called as such:
* NumMatrix obj = new NumMatrix(matrix);
* int param_1 = obj.sumRegion(row1,col1,row2,col2);
*/
Python:
class NumMatrix:
def __init__(self, matrix: List[List[int]]):
if not matrix or not matrix[0]:
return
self.dp = [[0] * (len(matrix[0]) + 1) for _ in range(len(matrix) + 1)]
for i in range(len(matrix)):
for j in range(len(matrix[i])):
self.dp[i + 1][j + 1] = self.dp[i][j + 1] + self.dp[i + 1][j] + matrix[i][j] - self.dp[i][j]
def sumRegion(self, row1: int, col1: int, row2: int, col2: int) -> int:
return self.dp[row2 + 1][col2 + 1] - self.dp[row1][col2 + 1] - self.dp[row2 + 1][col1] + self.dp[row1][col1];
# Your NumMatrix object will be instantiated and called as such:
# obj = NumMatrix(matrix)
# param_1 = obj.sumRegion(row1,col1,row2,col2)