LeetCode Maximal Square

Description:

Given a 2D binary matrix filled with 0's and 1's, find the largest square containing all 1's and return its area.

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Solution:

方法一：就一个个来咯，从小到大找，len从小到大，如果有len不存在，就返回最大值。

import java.util.*;

public class Solution {
public int maximalSquare(char[][] matrix) {
int m = matrix.length;
if (m == 0)
return 0;
int n = matrix[0].length;
int len = Math.min(m, n);

loop: for (int l = 1; l <= len; l++) {
for (int i = 0; i <= m - l; i++) {
for (int j = 0; j <= n - l; j++) {
if (matrix[i][j] == '1' && valid(matrix, i, j, l))
continue loop;
}
}
return (l - 1) * (l - 1);
}
return len * len;
}

boolean valid(char[][] matrix, int a, int b, int len) {
for (int i = a; i < a + len; i++)
for (int j = b; j < b + len; j++)
if (matrix[i][j] != '1')
return false;
return true;
}
}

方法二：也可以用DP来解决，dp[i][j]记录的是从i,j这个点开始，能走的最大的正方形的边长。所以如果dp[i][j],dp[i+1][j],dp[i][j+1]和dp[i+1][j+1]都等于l，则dp[i][j]=l+1

import java.util.*;

public class Solution {
public int maximalSquare(char[][] matrix) {
int m = matrix.length;
if (m == 0)
return 0;
int n = matrix[0].length;
int dp[][] = new int[m]
;
int max = 0;

for (int i = 0; i < m; i++)
for (int j = 0; j < n; j++)
if (matrix[i][j] == '1') {
dp[i][j] = 1;
max = 1;
}

int maxL = Math.max(m, n);
for (int l = 1; l < maxL; l++) {
for (int i = 0; i < m - l; i++)
for (int j = 0; j < n - l; j++) {
if (dp[i][j] == l && dp[i][j + 1] == l && dp[i + 1][j] == l
&& dp[i + 1][j + 1] == l) {
dp[i][j] = l + 1;
max = dp[i][j];
}
}
}

return max * max;
}
}

方法三：倒溯法，从最右下角来，如果dp[i][j] = min{dp[i+1][j], dp[i][j+1], dp[i+1][j+1] } +1

import java.util.*;

public class Solution {
public int maximalSquare(char[][] matrix) {
int m = matrix.length;
if (m == 0)
return 0;
int n = matrix[0].length;
int dp[][] = new int[m]
;
int max = 0;

for (int i = 0; i < m; i++)
if (matrix[i][n - 1] == '1') {
dp[i][n - 1] = 1;
max = 1;
}
for (int j = 0; j < n; j++)
if (matrix[m - 1][j] == '1') {
dp[m - 1][j] = 1;
max = 1;
}

for (int i = m - 2; i >= 0; i--)
for (int j = n - 2; j >= 0; j--)
if (matrix[i][j] == '1') {
dp[i][j] = Math.min(dp[i + 1][j + 1],
Math.min(dp[i + 1][j], dp[i][j + 1])) + 1;
max = Math.max(max, dp[i][j]);
}
return max * max;
}
}