LeetCode 85. Maximal Rectangle

原题链接在这里：https://leetcode.com/problems/maximal-rectangle/

题目：

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

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0

Return 6.

题解：

用数组heights存储一行中值为1的点的最大高度, 然后像Largest Rectangle in Histogram算这一行能产生的最大面积maxRec.

Time Complexity: O(m * n). m = matrix.length. n = matrix[0].length. Calculate area takes O(n) time, but it is within outer i loop, not inner j loop.

Space: O(n).

AC Java:

class Solution {
    public int maximalRectangle(char[][] matrix) {
        if(matrix == null || matrix.length == 0 || matrix[0].length == 0){
            return 0;
        }
        
        int m = matrix.length;
        int n = matrix[0].length;
        int [] heights = new int[n];
        int res = 0;
        
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                heights[j] = matrix[i][j] == '1' ? heights[j] + 1 : 0;
            }
            
            res = Math.max(res, calArea(heights));    
        }
        
        return res;
    }
    
    private int calArea(int [] heights){
        if(heights == null || heights.length == 0){
            return 0;
        }
        
        Stack<Integer> stk = new Stack<>();
        stk.push(-1);
        int res = 0;
        
        for(int i = 0; i < heights.length; i++){
            while(stk.peek() != -1 && heights[stk.peek()] >= heights[i]){
                int h = heights[stk.pop()];
                int w = i - stk.peek() - 1;
                res = Math.max(res, h * w);
            }
            
            stk.push(i);
        }
        
        while(stk.peek() != -1){
            int h = heights[stk.pop()];
            int w = heights.length - stk.peek() - 1;
            res = Math.max(res, h * w);
        }
        
        return res;
    }
}

类似Largest Rectangle in Histogram.