题目

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
• 1 <= row, cols <= 200
• matrix[i][j] 为 '0' 或 '1'

解答

源代码

class Solution {public int maximalRectangle(char[][] matrix) {int m = matrix.length, n = matrix[0].length;int[] heights = new int[n];int maxArea = 0;for (int i = 0; i < m; i++) {for (int j = 0; j < n; j++) {if (matrix[i][j] == '1') {heights[j] += 1;} else {heights[j] = 0;}}maxArea = Math.max(maxArea, largestRectangleArea(heights));}return maxArea;}public int largestRectangleArea(int[] heights) {int[] left = new int[heights.length];int[] right = new int[heights.length];Deque<Integer> stack = new ArrayDeque<>();for (int i = 0; i < heights.length; i++) {while (!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {stack.pop();}left[i] = stack.isEmpty() ? -1 : stack.peek();stack.push(i);}stack.clear();for (int i = heights.length - 1; i >= 0; i--) {while(!stack.isEmpty() && heights[stack.peek()] >= heights[i]) {stack.pop();}right[i] = stack.isEmpty() ? heights.length : stack.peek();stack.push(i);}int maxArea = 0;for (int i = 0; i < heights.length; i++) {maxArea = Math.max(maxArea, (right[i] - left[i] - 1) * heights[i]);}return maxArea;}
}

总结

这题是建立在「84.柱形图中最大的矩形」上进行解答的，最第一行到最后一行，把当前行到最上面看作一个柱形图，计算出每列的高度，利用 「84.柱形图中最大的矩形」的函数进行计算，最终比较得到最大值。