https://leetcode.com/problems/maximal-rectangle/
思路一:这道题和前面那道 84. Largest Rectangle in Histogram 类似,遍历二维数组,生成一个 heights 数组,然后用 Largest Rectangle in Histogram 一样的方法做。
public class Solution { public int maximalRectangle(char[][] matrix) { int result = 0; if (matrix == null || matrix.length == 0) { return result; } int[] heights = new int[matrix[0].length]; for (int i = 0; i < matrix.length; i++) { for (int j = 0; j < matrix[0].length; j++) { if (matrix[i][j] == '0') { heights[j] = 0; } else { heights[j] += 1; } } result = Math.max(result, helper(heights)); } return result; } private int helper(int[] heights) { int result = 0; Stack<Integer> stack = new Stack<>(); for (int i = 0; i < heights.length; i++) { while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) { int tmp = stack.pop(); int height = heights[tmp]; int width = stack.isEmpty() ? i : i - stack.peek() - 1; result = Math.max(result, height * width); } stack.push(i); } while (!stack.isEmpty() && heights[stack.peek()] >= 0) { int tmp = stack.pop(); int height = heights[tmp]; int width = stack.isEmpty() ? heights.length : heights.length - stack.peek() - 1; result = Math.max(result, height * width); } return result; } }
思路二:Dynamic Programming - https://discuss.leetcode.com/topic/6650/share-my-dp-solution