标签：sample highlight only public output int 最小值 binary ++i
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square which diagonal is all 1 and others is 0.
Only consider the main diagonal situation.
Example 1:
Input:
[[1,0,1,0,0],[1,0,0,1,0],[1,1,0,0,1],[1,0,0,1,0]]
Output:
9
Explanation:
[0,2]->[2,4]
Example 2:
Input: [[1,0,1,0,1],[1,0,0,1,1],[1,1,1,1,1],[1,0,0,1,0]] Output: 4 Explanation: [0,2]->[1,3]
思路：动态规划，u和l数组分别代表左边三角形的最大值和上方三角形的最大值，而f代表对角线到此点的最大长度。
直接三者求最小值转移即可。
public int maxSquare2(int[][] matrix) {
// write your code here
int n = matrix.length;
if (n == 0)
return 0;
int m = matrix[0].length;
if (m == 0)
return 0;
int[][] f = new int[n][m];
int[][] u = new int[n][m];
int[][] l = new int[n][m];
int length = 0;
for (int i = 0; i < n; ++i)
for (int j = 0; j < m; ++j) {
if (matrix[i][j] == 0) {
f[i][j] = 0;
u[i][j] = l[i][j] = 1;
if (i > 0)
u[i][j] = u[i - 1][j] + 1;
if (j > 0)
l[i][j] = l[i][j - 1] + 1;
} else {
u[i][j] = l[i][j] = 0;
if (i > 0 && j > 0)
f[i][j] = Math.min(f[i - 1][j - 1], Math.min(u[i - 1][j], l[i][j - 1])) + 1;
else
f[i][j] = 1;
}
length = Math.max(length, f[i][j]);
}
return length * length;
}
}
标签：sample highlight only public output int 最小值 binary ++i
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078292.html
