【leetcode刷题笔记】Maximal Rectangle

Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area.

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题解，很巧妙的一道题，对于一个0-1矩阵，它的每一行及以上都可以看作一个直方图（如下图所示），利用Largest Rectangle in Histogram的方法，可以在O(n)的时间搜索出这一行及以上的直方图中面积最大的矩形，对矩阵的每一行依次做这个操作，就可以在O(n²)的时间搜索出最大的矩形了。

将原矩阵转换成直方图矩阵的方法：设用transfer矩阵存放转换后的直方图，则

transfer[0][j] = matrix[0][j] - '0';

transfer[i][j] = (matrix[i][j] == '0' ? 0: transfer[i-1][j] + 1); (i >= 1)

代码如下：

public class Solution {
    public int largestRectangleArea(int[] height) {
        if(height == null || height.length == 0)
            return 0;
        Stack<Integer> index = new Stack<Integer>();
        int totalMax = 0;
        ArrayList<Integer> newHeight = new ArrayList<Integer>();
        for(int i:height) newHeight.add(i);
        newHeight.add(0);
        
        
        for(int i = 0;i < newHeight.size();i++){
            if(index.isEmpty() || newHeight.get(i) >= newHeight.get(index.peek()))
                index.push(i);
            else{
                int top = index.pop();
                totalMax = Math.max(totalMax,newHeight.get(top) * (index.isEmpty()?i:i-index.peek()-1));
                i--;
            }
        }
        
        return totalMax;
    }
    public int maximalRectangle(char[][] matrix) {
        if(matrix == null || matrix.length == 0)
            return 0;
        int m = matrix.length;
        int n = matrix[0].length;
        int totalMax = 0;
        int[][] transfer = new int[m][n];
        
        //transform matrix to histogram in a new matrix
        for(int i = 0;i < n;i++)
            transfer[0][i] = matrix[0][i]- '0';
        for(int i = 1;i < m;i++){
            for(int j = 0;j < n;j++){
                if(matrix[i][j] == '0' )
                    transfer[i][j] = 0;
                else {
                    transfer[i][j] = transfer[i-1][j] + 1; 
                }
            }
        }
        
        //Using histogram to find the biggest rectangle
        for(int i = 0;i < m;i++){
            totalMax = Math.max(totalMax, largestRectangleArea(transfer[i]));
        }
        
        return totalMax;
    }
}