221.Maximal-Square

221. Maximal Square

题目地址

https://leetcode.com/problems/maximal-square/

题目描述

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

Example:

Input: 
1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Output: 4

代码

Approach #1 Brute Force

Time complexity : O((m*n)^2)

class Solution {
  public int maximalSquare(char[][] matrix) {
        int rows = matrix.length, cols = rows > 0 ? matrix[0].length: 0;
    int maxsqlen = 0;
    for (int i = 0; i < rows; i++) {
      for (int j = 0; j < cols; j++) {
        if (matrix[i][j] == '1') {
          int sqlen = 1;
          boolean flag = true;
          while (sqlen + i < rows && sqlen + j < cols && flag) {
            for (int k = j; k <= sqlen + j; k++) {
              if (matrix[i + sqlen][k] == '0') {
                flag = false;
                break;
              }
            }
            for (int k = i; k <= sqlen + i; k++) {
              if (matrix[k][j + sqlen] == '0') {
                flag = false;
                break;
              }
            }
            if (flag)        sqlen++;
          }
          if (maxsqlen < sqlen) {
            maxsqlen = sqlen;
          }
        }
      }
    }

    return maxsqlen * maxsqlen;
  }
}

Approach #2 Dynamic Programming

class Solution {
     public int maximalSquare(char[][] matrix) {
    int rows = matrix.length;
    int cols = rows > 0 ? matrix[0].length : 0;
    int[][] dp = new int[rows + 1][cols + 1];
    int maxsqlen = 0;
    for (int i = 1; i <= rows; i++) {
      for (int j = 1; j <= cols; j++) {
        if (matrix[i-1][j-1] == '1') {
          dp[i][j] = Math.min(Math.min(dp[i][j - 1], dp[i - 1][j]), dp[i-1][j-1]) + 1;
            maxsqlen = Math.max(maxsqlen, dp[i][j]);
        }
      }
    }

    return maxsqlen * maxsqlen;
  }
}

Approach #3 Better Dynamic Programming

dp[j] = min(dp[j - 1], dp[j], prev)

public class Solution {
    public int maximalSquare(char[][] matrix) {
        int rows = matrix.length, cols = rows > 0 ? matrix[0].length : 0;
        int[] dp = new int[cols + 1];
        int maxsqlen = 0, prev = 0;
        for (int i = 1; i <= rows; i++) {
            for (int j = 1; j <= cols; j++) {
                int temp = dp[j];
                if (matrix[i - 1][j - 1] == '1') {
                    dp[j] = Math.min(Math.min(dp[j - 1], prev), dp[j]) + 1;
                    maxsqlen = Math.max(maxsqlen, dp[j]);
                } else {
                    dp[j] = 0;
                }
                prev = temp;
            }
        }
        return maxsqlen * maxsqlen;
    }
}