标签：情况 color 计算 bsp 矩阵 where 空间复杂度 cto div
Given a m * n matrix mat and an integer K, return a matrix answer where each answer[i][j] is the sum of all elements mat[r][c] for i - K <= r <= i + K, j - K <= c <= j + K, and (r, c) is a valid position in the matrix.
Example 1:
Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 1
Output: [[12,21,16],[27,45,33],[24,39,28]]
Example 2:
Input: mat = [[1,2,3],[4,5,6],[7,8,9]], K = 2
Output: [[45,45,45],[45,45,45],[45,45,45]]
Constraints:
m == mat.lengthn == mat[i].length1 <= m, n, K <= 1001 <= mat[i][j] <= 100题目大意：
给你一个 m * n 的矩阵 mat 和一个整数 K ，请你返回一个矩阵 answer ，其中每个 answer[i][j] 是所有满足下述条件的元素 mat[r][c] 的和：
i - K <= r <= i + K, j - K <= c <= j + K
(r, c) 在矩阵内。
思路：这题其实可以看作对灰度图做均值滤波，滤波器大小为$K \times K$。如果直接计算的话，时间复杂度为$O(m \times n \times K \times K)$。采用动态规划思想，初始化二维矩阵dp，dp[i][j]表示矩阵mat从(0,0)到(i,j)组成的矩阵之和。dp[i][j] = dp[i - 1][j] + dp[i][j - 1] - dp[i - 1][j - 1] + mat[i][j],同时考虑i=0或者j=0的边界情况。
当计算answer[i][j]时，通常情况下answer[i][j] = dp[i + K][j + K] - dp[i + K][j - K - 1] - dp[i - K - 1][j + K] + dp[i - K - 1][j - K - 1]。同时考虑i + K >= row, j + k >= col, i - K - 1 < 0, j - K - 1 < 0的边界情况就好了。
1 class Solution { 2 public: 3 vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int K) { 4 int row = mat.size(), col = (row > 0) ? mat[0].size() : 0; 5 vector<vector<int>> dp(row, vector<int>(col, 0)); 6 dp[0][0] = mat[0][0]; 7 for(int i = 1; i < row; ++i) { 8 dp[i][0] = dp[i - 1][0] + mat[i][0]; 9 } 10 for (int j = 1; j < col; ++j) { 11 dp[0][j] = dp[0][j - 1] + mat[0][j]; 12 } 13 for (int i = 1; i < row; ++i) { 14 for (int j = 1; j < col; ++j) { 15 dp[i][j] = dp[i][j - 1] + dp[i - 1][j] - dp[i - 1][j - 1] + mat[i][j]; 16 } 17 } 18 vector<vector<int>> ans(row, vector<int>(col, 0)); 19 for (int i = 0; i < row; ++i) { 20 for (int j = 0; j < col; ++j) { 21 int left = (j - K - 1) >= 0 ? dp[min(i + K, row - 1)][j - K - 1] : 0; 22 int top = (i - K - 1) >= 0 ? dp[i - K - 1][min(j + K, col - 1)] : 0; 23 int topleft = (i - K - 1) >= 0 && (j - K - 1) >= 0 ? dp[i - K - 1][j - K - 1] : 0; 24 ans[i][j] = dp[min(i + K, row - 1)][min(j + K, col - 1)] - left - top + topleft; 25 } 26 } 27 return ans; 28 } 29 };
时间复杂度：$O(m \times n)$
空间复杂度：$O(m \times n)$
leetcode 1314. Matrix Block Sum
标签：情况 color 计算 bsp 矩阵 where 空间复杂度 cto div
原文地址：https://www.cnblogs.com/qinduanyinghua/p/13047110.html
