标签：style blog io ar color os sp for on
Given a 2D binary matrix filled with 0‘s and 1‘s, find the largest rectangle containing all ones and return its area.
Analysis:
For each position (i,j), we calculate how many accumlative 1s at the left direction of this position at this row. We record this number at d[i][j]. For example, for matrix:
0 0 0 1 1 1
0 1 1 1 1 1
0 1 1 1 1 1
1 0 1 1 1 1
We have matrix d[i][j]
0 0 0 1 2 3
0 1 2 3 4 5
0 1 2 3 4 5
1 0 1 2 3 4
Then, at each postion (i,j), we scan the matrix along the upper direction and calculate the area of the maximum 1 matrix with the right bottom corner at (i,j). For row k<i, the 1s matrix should be (k-i+1)*min(d[l][j], l=k...i);
After caluclating the 1s matrix for each (i,j), we get the answer.
Solution:
1 public class Solution { 2 public int maximalRectangle(char[][] matrix) { 3 int rowNum = matrix.length; 4 if (rowNum==0) return 0; 5 int colNum = matrix[0].length; 6 if (colNum==0) return 0; 7 8 int[][] d = new int[rowNum][colNum]; 9 for (int i=0;i<rowNum;i++) 10 for (int j=0;j<colNum;j++) 11 if (j==0) 12 if (matrix[i][j]==‘1‘) d[i][j]=1; 13 else d[i][j]=0; 14 else 15 if (matrix[i][j]==‘1‘) d[i][j]=d[i][j-1]+1; 16 else d[i][j]=0; 17 18 int maxSoFar = -1; 19 for (int i=0;i<rowNum;i++) 20 for (int j=0;j<colNum;j++){ 21 int curMax = d[i][j]; 22 int minLen = d[i][j]; 23 for (int k=(i-1);k>=0;k--){ 24 if (d[k][j]==0) break; 25 if (d[k][j]<minLen) minLen = d[k][j]; 26 int newMax = (i-k+1)*minLen; 27 if (curMax<newMax) curMax = newMax; 28 } 29 if (curMax>maxSoFar) maxSoFar = curMax; 30 31 } 32 33 return maxSoFar; 34 } 35 }
标签：style blog io ar color os sp for on
原文地址：http://www.cnblogs.com/lishiblog/p/4117867.html
