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给定一个仅包含 0 和 1 、大小为 rows x cols 的二维二进制矩阵,找出只包含 1 的最大矩形,并返回其面积。
示例 1: 输入:matrix = [["1","0","1","0","0"],["1","0","1","1","1"],["1","1","1","1","1"],["1","0","0","1","0"]] 输出:6 解释:最大矩形如上图所示。 示例 2: 输入:matrix = [] 输出:0 示例 3: 输入:matrix = [["0"]] 输出:0 示例 4: 输入:matrix = [["1"]] 输出:1 示例 5: 输入:matrix = [["0","0"]] 输出:0 提示: rows == matrix.length cols == matrix[0].length 0 <= row, cols <= 200 matrix[i][j] 为 '0' 或 '1'
class Solution { public int maximalRectangle(char[][] matrix) { int rows = matrix.length; if(rows == 0) return 0; int columns = matrix[0].length; int[][] left = new int[rows][columns]; //寻找每一行横着看的 左边有几个连续的 for(int i=0;i<rows;i++){ for(int j=0;j<columns;j++){ if(matrix[i][j] == '1'){ left[i][j] = (j == 0 ? 0:left[i][j-1]) + 1; } } } int res = 0; for(int i = 0;i < rows; i++){ for(int j = 0; j < columns; j++){ if(matrix[i][j] == '1'){ //高度为1的 int width = left[i][j]; int area = width; //高度为2,3,4,5...... for(int h = i - 1;h >= 0; h--){ width = Math.min(width, left[h][j]); area = Math.max(area, width * (i - h + 1)); } res = Math.max(res, area); } } } return res; } }
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