标签：min code find down htm eset sub ret 规划
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest square containing all 1‘s and return its area.
Example 1:
Input:
[
[1, 0, 1, 0, 0],
[1, 0, 1, 1, 1],
[1, 1, 1, 1, 1],
[1, 0, 0, 1, 0]
]
Output: 4
Example 2:
Input: [ [0, 0, 0], [1, 1, 1] ] Output: 1
思路：
可以归类为动态规划题目, 推荐用递推来实现.
设定状态: dp[i][j] 表示以(i, j)为右下顶点的最大全1矩阵的边长.
状态转移方程:
if matrix[i][j] == 0
dp[i][j] = 0
else // 此时为dp[i-1][j-1], dp[i-1][j], dp[i][j-1] 确定的区域的最大全1矩阵
dp[i][j] = min{dp[i-1][j-1], dp[i-1][j], dp[i][j-1]} + 1 // 得到此方程需要一定推导, 纸笔画一下
边界: if i == 0 or j == 0: dp[i][j] = matrix[i][j]
答案: max\{dp[i][j]\}^2max{dp[i][j]}2
public class Solution {
/**
* @param matrix: a matrix of 0 and 1
* @return: an integer
*/
public int maxSquare(int[][] matrix) {
// write your code here
int ans = 0;
int n = matrix.length;
int m;
if (n > 0)
m = matrix[0].length;
else
return ans;
int [][]res = new int [n][m];
for (int i = 0; i < n; i++) {
res[i][0] = matrix[i][0];
ans = Math.max(res[i][0] , ans);
for (int j = 1; j < m; j++) {
if (i > 0) {
if (matrix[i][j] > 0) {
res[i][j] = Math.min(res[i-1][j], Math.min(res[i][j-1], res[i-1][j-1])) + 1;
} else {
res[i][j] = 0;
}
} else {
res[i][j] = matrix[i][j];
}
ans = Math.max(res[i][j], ans);
}
}
return ans * ans;
}
}
标签：min code find down htm eset sub ret 规划
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078212.html
