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.length​
  • ​n == mat[i].length​
  • ​1 <= m, n, K <= 100​
  • ​1 <= mat[i][j] <= 100​

1.Rock大神



class Solution {
public int[][] matrixBlockSum(int[][] mat, int K) {
int m = mat.length, n = mat[0].length;
int[][] rangeSum = new int[m + 1][n + 1];
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j)
rangeSum[i + 1][j + 1] = rangeSum[i + 1][j] + rangeSum[i][j + 1] - rangeSum[i][j] + mat[i][j];
int[][] ans = new int[m][n];
for (int i = 0; i < m; ++i)
for (int j = 0; j < n; ++j) {
int r1 = Math.max(0, i - K), c1 = Math.max(0, j - K), r2 = Math.min(m, i + K + 1), c2 = Math.min(n, j + K + 1);
ans[i][j] = rangeSum[r2][c2] - rangeSum[r2][c1] - rangeSum[r1][c2] + rangeSum[r1][c1];
}
return ans;
}
}


1314. Matrix Block Sum_java

 

 ​​https://leetcode.com/problems/matrix-block-sum/discuss/477036/JavaPython-3-PrefixRange-sum-w-analysis-similar-to-LC-30478​

2. 下面这个三哥的也可以

​https://leetcode.com/problems/matrix-block-sum/discuss/477041/Java-Prefix-sum-with-Picture-explain-Clean-code-O(m*n)​



class Solution {
public int[][] matrixBlockSum(int[][] mat, int K) {
int m = mat.length, n = mat[0].length;
int[][] sum = new int[m + 1][n + 1]; // sum[i][j] is sum of all elements from rectangle (0,0,i,j) as left, top, right, bottom corresponding
for (int r = 1; r <= m; r++) {
for (int c = 1; c <= n; c++) {
sum[r][c] = mat[r - 1][c - 1] + sum[r - 1][c] + sum[r][c - 1] - sum[r - 1][c - 1];
}
}
int[][] ans = new int[m][n];
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
int r1 = Math.max(0, r - K), c1 = Math.max(0, c - K);
int r2 = Math.min(m - 1, r + K), c2 = Math.min(n - 1, c + K);
r1++; c1++; r2++; c2++; // Since `sum` start with 1 so we need to increase r1, c1, r2, c2 by 1
ans[r][c] = sum[r2][c2] - sum[r2][c1-1] - sum[r1-1][c2] + sum[r1-1][c1-1];
}
}
return ans;
}
}


3. 最后总结出来,此题和304一毛一样几乎



class Solution {
private int[][] sum;
public int[][] matrixBlockSum(int[][] mat, int K) {
int m = mat.length, n = mat[0].length;
sum = new int[m + 1][n + 1];
for (int r = 1; r <= m; r++) {
for (int c = 1; c <= n; c++) {
sum[r][c] = mat[r - 1][c - 1] + sum[r - 1][c] + sum[r][c - 1] - sum[r - 1][c - 1];
}
}
int[][] ans = new int[m][n];
for (int r = 0; r < m; r++) {
for (int c = 0; c < n; c++) {
int r1 = Math.max(0, r - K), c1 = Math.max(0, c - K);
int r2 = Math.min(m - 1, r + K), c2 = Math.min(n - 1, c + K);
ans[r][c] = sumRegion(r1,c1,r2,c2);
}
}
return ans;
}
public int sumRegion(int row1, int col1, int row2, int col2) {
return sum[row2 + 1][col2 + 1] - sum[row2 + 1][col1]
- sum[row1][col2 + 1] + sum[row1][col1];
}
}


 

第一步,得到range sum(整体的dp)

1314. Matrix Block Sum_java_02

 

 

第二步,得到block sum

1314. Matrix Block Sum_dp_03

 为什么是这样?参考下面的图

 1314. Matrix Block Sum_java_04

 

1314. Matrix Block Sum_it_05

 

 

 

Reference:304. range sum query

一定要注意下标!!!!!!!!!!!!!!!!!!!!!!!!!

一定要注意下标!!!!!!!!!!!!!!!!!!!!!!!!!

r1, c1, r2, c2和我们要求的不一定相同(一定不相同!!!!!!!!!!!

上限和下限