LeetCode刷题：85. Maximal Rectangle

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LeetCode刷题：85. Maximal Rectangle

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

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

Example:

Input:
[
  ["1","0","1","0","0"],
  ["1","0","1","1","1"],
  ["1","1","1","1","1"],
  ["1","0","0","1","0"]
]
Output: 6

题目的大意是：在一个给定的二维矩阵中，数组元素由0或者1组成，要找出一个由1构成的最大矩形使得面积最大。

在解答这道题目是，网上很多解法都参考了 LeetCode 84 题的解法。

LeetCode 84. Largest Rectangle in Histogram https://leetcode.com/problems/largest-rectangle-in-histogram/

解题：https://blog.csdn.net/seagal890/article/details/89074836

本题的解法：

    // 首先，也是用dp做：heights保存某一列从上数第i行的连续的'1'的个数；
    // 然后使用84的逻辑，取连续区间最大的矩形大小就可以了；
    public int maximalRectangle(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0] == null || matrix[0].length == 0) return 0;
        int width = matrix[0].length, res = 0;
        int[] heights = new int[width];
        for (char[] row : matrix) {
            for (int c = 0; c < width; c++) {
                if (row[c] == '1') heights[c]++;
                else heights[c] = 0;
            }
            res = Math.max(res, largestRectangleArea(heights));
        }
        return res;
    }
    
	// 以下部分是第84题的解法：分治求最大矩形（不需要修改）
    // 思路：对于每一段区间，都存在一个最小值
		// 对于最小值，无非就是三种可能：
		// 1：要么整段面积最大，2、3：要么最小值左边或者最小值右边（均不包含最小值）的面积最大，采用分治法递归解决；
		// 遇到有序排列的区间，采用递归会降低效率，于是只要单独计算并且比较就可以
	public int largestRectangleArea(int[] heights) {
        return largestRect(heights, 0, heights.length - 1);
    }
	
    private int largestRect(int[] heights, int start, int end) {
        if (start > end) return 0;
        if (start == end) return heights[start];
        int minIndex = start;
        // 使用是否有序排列的变量可以显著提高效率
        // 这里可以检测双向（变大或者变小的顺序）
        int inc = 0, dec = 0;
        for (int i = start + 1; i <= end; i++) {
            if (heights[i] < heights[minIndex]) minIndex = i;
            if (heights[i] > heights[i - 1]) inc++; // 升序
            else if (heights[i] < heights[i - 1]) dec--; // 降序
        }
        int res = 0;
        // 升序
        if (dec == 0) {
            for (int i = start; i <= end; i++)
                res = Math.max(res, heights[i] * (end - i + 1));
        } // 降序
        else if (inc == 0) {
            for (int i = start; i <= end; i++)
                res = Math.max(res, heights[i] * (i - start + 1));
        } // 无序
        else {
            res = Math.max(Math.max(largestRect(heights, minIndex + 1, end), largestRect(heights, start, minIndex - 1)),
                    heights[minIndex] * (end - start + 1));
        }
        return res;
    }

另外一种解法：

public int MaximalRectangle(char[,] matrix) {
        // Heights accum
        int[] dp = new int[matrix.GetLength(1)];
        
        int max = 0;
        
        for(int i = 0; i < matrix.GetLength(0); ++i)
        {
	    // Accumlate heights for current row
            for(int j = 0; j < matrix.GetLength(1); ++j)
            {                
                dp[j] = matrix[i,j] == '1'? dp[j] + 1 : 0;
            }
            
	    // Finding maximum rectangle in current state, O(columns)
            max = Math.Max(max, LargestRectangleArea(dp));
        }
        
        return max;
        
    }
    
    // Largest rectangle in historgram, leet code #84
    int LargestRectangleArea(int[] heights)
    {

        Stack<int> st = new Stack<int>();

        int left = 0;
        int max = 0;

        while (left <= heights.Length)
        {
            while (st.Count > 0 && (left == heights.Length || heights[st.Peek()] > heights[left]))
            {
                var h = heights[st.Pop()];
                if (st.Count == 0) max = Math.Max(max, h * left);
                else max = Math.Max(max, h * (left - st.Peek() - 1));
            }
            st.Push(left++);
        }
        
        return max;
        
    }