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topic
matrix area and
Given a m x n matrix mat and an integer k, return a matrix answer where each answer[i][j] is the sum of all elements that satisfy the following conditions mat[r][c]:
i - k <= r <= i + k,
j - k <= c <= j + kand
(r, c) in the matrix.
Example 1:
输入：mat = [[1,2,3],[4,5,6],[7,8,9]], k = 1
输出：[[12,21,16],[27,45,33],[24,39,28]]
复制代码Example 2:
输入：mat = [[1,2,3],[4,5,6],[7,8,9]], k = 2
输出：[[45,45,45],[45,45,45],[45,45,45]]
复制代码hint:
m == mat.length
n == mat[i].length
1 <= m, n, k <= 100
1 <= mat[i][j] <= 100
复制代码answer
problem-solving analysis
Problem solving ideas
- This problem is a typical dynamic programming problem;
- dp[i][j] represents the maximum square side length that can be formed by the lower right corner of the i-th row and the j-th column, then the recursion formula is:
dp[i][j] = 1 + min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]);
复制代码Then max * maxget the maximum value of the rectangular area by .
3. The problem-solving code is as follows:
Complexity
Time Complexity: O(M * N)
Space Complexity:O(M * N)
problem solving code
The solution code is as follows (detailed comments in the code):
class Solution {
public int maximalSquare(char[][] matrix) {
/**
dp[i][j]表示以第i行第j列为右下角所能构成的最大正方形边长, 则递推式为:
dp[i][j] = 1 + min(dp[i-1][j-1], dp[i-1][j], dp[i][j-1]);
**/
int m = matrix.length;
if(m < 1) return 0;
int n = matrix[0].length;
int max = 0;
int[][] dp = new int[m+1][n+1];
for(int i = 1; i <= m; ++i) {
for(int j = 1; j <= n; ++j) {
if(matrix[i-1][j-1] == '1') {
dp[i][j] = 1 + Math.min(dp[i-1][j-1], Math.min(dp[i-1][j], dp[i][j-1]));
max = Math.max(max, dp[i][j]);
}
}
}
return max*max;
}
}
复制代码Feedback results after submission (because this topic has not been optimized, the performance is average):