最大矩形
难度：困难
题目描述
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
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
题解
是0084的进阶，可以将它升维转换为二维数组，具体如下
比如matrix = [[“1”,“0”,“1”,“0”,“0”],[“1”,“0”,“1”,“1”,“1”],[“1”,“1”,“1”,“1”,“1”],[“1”,“0”,“0”,“1”,“0”]]
可以转换为：
4 0 3 0 0
3 0 2 3 2
2 1 1 2 1
1 0 0 1 0
即：将需要讨论的内容缩小，将本来需要讨论的纵坐标转化为每一行的高，之后就可以根据0084的算法来对每一行的内容都取最大面积，之后对结果进行对比，最大的数即为结果
想法代码
class Solution
{public static void Main(String[] args){char[][] matrix = {new[] { '1', '0', '1', '0', '0' },new[] { '1', '0', '1', '1', '1' },new[] { '1', '1', '1', '1', '1' },new[] { '1', '0', '0', '1', '0' }};Solution solution = new Solution();int ans = solution.MaximalRectangle(matrix);Console.WriteLine(ans);}public int MaximalRectangle(char[][] matrix){int maxArea = 0;int rows = matrix.Length;int cols = matrix[0].Length;int[] heights = new int[cols];for (int i = 0; i < rows; i++){for (int j = 0; j < cols; j++){if (matrix[i][j] == '0'){heights[j] = 0;}else{heights[j]++;}}maxArea = Math.Max(maxArea, GetMaxArea(heights));}return maxArea;}public int GetMaxArea(int[] heights){int[] left = new int[heights.Length];int[] right = new int[heights.Length];Array.Fill(left, -1);Array.Fill(right, heights.Length);Stack<int> stack = new Stack<int>();for (int i = 0; i < heights.Length; i++){int height = heights[i];while (stack.Count > 0 && heights[stack.Peek()] >= height){right[stack.Pop()] = i;}if (stack.Count > 0){left[i] = stack.Peek();}stack.Push(i);}int ans = 0;for (int i = 0; i < heights.Length; i++){ans = Math.Max(ans, (right[i] - left[i] - 1) * heights[i]);}return ans;}
}
