【LeetCode从零单排】No221.Maximal Square

题目

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.

解法很巧妙，我也是看了discuss，首先列出最左一列和最上面一列，剩下的解法可以看这个

Basic idea is to iterate over all columns and rows of a matrix (starting with i=j=1). If value in a cell>0 and cells to the north, west, and north-west are >0, pick smallest value of those 3 cells, take it's square root, add 1, and assign square of new value to current cell. For example given matrix

1   1   1   1   1   1
1   1   1   0   1   1
1   1   1   1   1   1
1   1   1   1   1   1
1   1   1   1   1   1
1   1   1   1   0   1

We get:

1   1   1   1   1   1
1   4   4   0   1   4
1   4   9   1   1   4
1   4   9   4   4   4
1   4   9   9   9   9
1   4   9  16   0   1

Our answer is the largest value in new matrix: 16

代码

public class Solution {
    public int maximalSquare(char[][] matrix) {
        if(matrix == null || matrix.length == 0) {
            return 0;
        }

        int res = 0;
        int m = matrix.length;
        int n = matrix[0].length;
        int[][] sq = new int[m][n];

        for(int i = 0; i < m; i++) {
            if(matrix[i][0] == '1') {
                sq[i][0] = 1;
                res = 1;
            } else {
                sq[i][0] = 0;
            }
        }
        for(int j = 0; j < n; j++) {
            if(matrix[0][j] == '1') {
                sq[0][j] = 1;
                res = 1;
            } else {
                sq[0][j] = 0;
            }
        }

        for(int i = 1; i < m; i++) {
            for(int j = 1; j < n; j++) {
               if(matrix[i][j] == '1') {
                   int min = Math.min(Math.min(sq[i][j-1], sq[i-1][j]), sq[i-1][j-1]);
                   //if(min != 0) {
                       sq[i][j] = (int)Math.pow(Math.sqrt(min)+1, 2);
                       res = Math.max(res, sq[i][j]);
                   //}
               }
            }
        }

        return res;
    }
} 

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