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  • leetcode 221: Maximal Square

    Maximal Square

    Total Accepted: 1312 Total Submissions: 6388

    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
    

    Return 4.

    [思路]

    dynamic programing.  以当前点(x,y) = '1' 为右下角的最大正方形的边长f(x,y) = min( f(x-1,y), f(x,y-1), f(x-1,y-1)) + 1.

    递推公式已建立, dp就自然而然了.

    [CODE]

    public class Solution {
        public int maximalSquare(char[][] matrix) {
            if(matrix==null || matrix.length==0 || matrix[0].length==0) return 0;
            
            int n = matrix.length;
            int m = matrix[0].length;
            
            int[][] d = new int[n][m];
            int max = 0;
    
            for(int i=0; i<n; i++) {
                if(matrix[i][0]=='1') {
                    d[i][0] = 1;
                    max = 1;
                }
            }
            
            for(int j=0; j<m; j++) {
                if(matrix[0][j]=='1') {
                    d[0][j] = 1;
                    max = 1;
                }
            }
            for(int i=1; i<n; i++) {
                for(int j=1; j<m; j++) {
                    if(matrix[i][j]=='0') d[i][j]=0;
                    else {
                        d[i][j] = Math.min( Math.min( d[i-1][j], d[i][j-1]), d[i-1][j-1] ) + 1;
                        max = Math.max(max, d[i][j]);
                    }
                }
            }
            return max*max;
        }
    }


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  • 原文地址:https://www.cnblogs.com/clnchanpin/p/7081655.html
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