給你一個 m * n 的矩陣 mat 和一個整數 K ,請你返回一個矩陣 answer ,其中每個 answer[i][j] 是所有滿足下述條件的元素 mat[r][c] 的和:
i - K <= r <= i + K, j - K <= c <= j + K
(r, c) 在矩陣內。
示例 1:
輸入:mat = [[1,2,3],[4,5,6],[7,8,9]], K = 1
輸出:[[12,21,16],[27,45,33],[24,39,28]]
示例 2:
輸入:mat = [[1,2,3],[4,5,6],[7,8,9]], K = 2
輸出:[[45,45,45],[45,45,45],[45,45,45]]
提示:
m == mat.length
n == mat[i].length
1 <= m, n, K <= 100
1 <= mat[i][j] <= 100
來源:力扣(LeetCode)
鏈接:https://leetcode-cn.com/problems/matrix-block-sum
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class Solution {
public:
vector<vector<int>> matrixBlockSum(vector<vector<int>>& mat, int K) {
int sum[110][110];
int n = mat.size(), m = mat[0].size();
memset(sum, 0, 0);
for(int i = 1; i <= n; i++)
for(int j = 1; j <= m; j++)
sum[i][j] = mat[i - 1][j - 1] + sum[i - 1][j] + sum[i][j - 1] - sum[i - 1][ j - 1 ];
vector<vector<int>> ret(n, vector<int>(m));
for(int i = 0; i < n; i++)
for(int j = 0; j < m; j++)
{
int a = i + 1, b = j + 1;
int x4 = min(a + K, n), y4 = min(b + K, m);
int x1 = max(0, a - K - 1), y1 = max(0, b - K - 1);
int x2 = max(0, a - K - 1), y2 = min(b + K, m);
int x3 = min(n, a + K), y3 = max(0, b - K - 1);
ret[i][j] = sum[x4][y4] - sum[x2][y2] - sum[x3][y3] + sum[x1][y1];
}
return ret;
}
};