Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing all ones and return its area.
public class Solution { public int maximalRectangle(char[][] matrix) { int row = matrix.length; if(row == 0) return 0; int column = matrix[0].length; int[][] consecutiveOnes = new int[row][column]; for(int i = 0; i < row; ++i) { for(int j = 0; j < column; ++j){ if(matrix[i][j] - ‘0‘ == 1){ if(j == 0) consecutiveOnes[i][j] = 1; else consecutiveOnes[i][j] = consecutiveOnes[i][j - 1] + 1; } else consecutiveOnes[i][j] = 0; } } //calculate the max area. we check each element of consecutiveOnes column by column int maxArea = 0; for(int i = 0; i < column; ++i){ for(int j = 0; j < row; ++j){ if(consecutiveOnes[j][i] != 0){ int minWidth = consecutiveOnes[j][i]; int index = j; while(index >= 0 && consecutiveOnes[index][i] != 0){ minWidth = Math.min(minWidth, consecutiveOnes[index][i]); maxArea = Math.max(maxArea, minWidth * (j - index + 1)); --index; } } } } return maxArea; } }
Remark: the second part (to seek the max area) can be solved by method used in "largest rectangle histogram" problem. Therefore, there is an algorithm with time O(row * column).
leetcode--Maximal Rectangle,布布扣,bubuko.com
原文地址:http://www.cnblogs.com/averillzheng/p/3825713.html