LeetCode Maximal Square

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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 01 0 1 1 11 1 1 1 11 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][n];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][n];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;}}

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