題目

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，剩下的解法可以看這個(gè)

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 1We 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

代碼

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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* 本文來自博客 ?“李博Garvin“

* 轉(zhuǎn)載請(qǐng)標(biāo)明出處:http://blog.csdn.net/buptgshengod

******************************************/

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