【LeetCode】Maximal Rectangle && Maximal Square

1、Maximal Rectangle 
Total Accepted: 24875 Total Submissions: 113379 My Submissions Question Solution 
Given a 2D binary matrix filled with 0's and 1's, 
find the largest rectangle containing all ones and return its area.
2、Maximal Square 
Total Accepted: 1488 Total Submissions: 7283 My Submissions Question Solution 
Given a 2D binary matrix filled with 0's and 1's, 
find the largest square containing all 1's and return its area.
For example, given the following matrix:

1 0 1 0 01 0 1 1 11 1 1 1 11 0 0 1 0

Return 4.
Credits:
Special thanks to @Freezen for adding this problem and creating all test cases.

【解题思路】
参考Largest Rectangle in Histogram，解题思路参考【LeetCode】Largest Rectangle in Histogram，
针对二维矩阵的每行，其实就是一个单独的Largest Rectangle。
针对长方形，思路和Largest Rectangle一致。
针对正方形，每次计算square的时候，取长和宽最小值的平方即为面积。

1 Java AC

public class Solution {    public int maximalRectangle(char[][] matrix) {        if(matrix == null || matrix.length == 0){            return 0;        }        int m = matrix.length;        int n = matrix[0].length;        int height[][] = new int[m][n];        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(matrix[i][j] == '0'){                    height[i][j] = 0;                }else{                    height[i][j] = i == 0 ? 1 : height[i-1][j] + 1;                }            }        }        int maxArea = 0;        for(int i = 0; i < m; i++){            int area = largestRectangleArea(height[i]);            maxArea = area > maxArea ? area : maxArea;        }        return maxArea;    }        public int largestRectangleArea(int[] height) {        if(height == null || height.length == 0){            return 0;        }        int len = height.length;        Stack<Integer> stack = new Stack<Integer>();        int area = 0;        for(int i = 0; i <= len; i++){            int h = i == len ? 0 : height[i];            if(stack.isEmpty() || height[stack.peek()] <= h){                stack.push(i);                continue;            }            int start = stack.pop();            int width = stack.empty() ? i : i - stack.peek() - 1;              area = Math.max(area, height[start] * width);              i--;          }        return area;    }}

2 Java AC

public class Solution {    public int maximalSquare(char[][] matrix) {        if(matrix == null || matrix.length == 0){            return 0;        }        int m = matrix.length;        int n = matrix[0].length;        int height[][] = new int[m][n];        for(int i = 0; i < m; i++){            for(int j = 0; j < n; j++){                if(matrix[i][j] == '0'){                    height[i][j] = 0;                }else{                    height[i][j] = i == 0 ? 1 : height[i-1][j] + 1;                }            }        }        int maxArea = 0;        for(int i = 0; i < m; i++){            int area = largestRectangleArea(height[i]);            maxArea = area > maxArea ? area : maxArea;        }        return maxArea;    }        public int largestRectangleArea(int[] height) {        if(height == null || height.length == 0){            return 0;        }        int len = height.length;        Stack<Integer> stack = new Stack<Integer>();        int area = 0;        for(int i = 0; i <= len; i++){            int h = i == len ? 0 : height[i];            if(stack.isEmpty() || height[stack.peek()] <= h){                stack.push(i);                continue;            }            int start = stack.pop();            int width = stack.empty() ? i : i - stack.peek() - 1;            int min = Math.min(height[start], width);            area = Math.max(area, min*min);              i--;          }        return area;    }}