原题传送门

1. 题目描述

2. Solution 1

1、思路分析
以 example 1. 为例
matrix = [
["1","0","1","0","0"],
["1","0","1","1","1"],
["1","1","1","1","1"],
["1","0","0","1","0"]
]

分析: 遍历 行(row)
row = 1: 可以得到 heights = [1, 0, 1, 0, 0] 套用 Q84
row = 2: -> heights = [2, 0, 2, 1, 1]
......

2、代码实现

package Q0099.Q0085MaximalRectangle;

import java.util.ArrayDeque;
import java.util.Deque;

/*
   以 example 1. 为例
    matrix = [
      ["1","0","1","0","0"],
      ["1","0","1","1","1"],
      ["1","1","1","1","1"],
      ["1","0","0","1","0"]
    ]

    分析: 遍历 行(row)
    row = 1: 可以得到 heights = [1, 0, 1, 0, 0] 套用 Q84
    row = 2: -> heights = [2, 0, 2, 1, 1]
    ......
  */
public class Solution1 {
    public int maximalRectangle(char[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return 0;
        int col = matrix[0].length;
        int[] heights = new int[col];
        int max = 0;
        for (char[] chars : matrix) {
            for (int j = 0; j < col; j++) {
                if (chars[j] == '1') {
                    heights[j]++;
                } else {
                    heights[j] = 0;
                }
            }
            int area = largestRectangleArea(heights);
            max = Math.max(max, area);
        }
        return max;
    }

    // 来自 Q84
    public int largestRectangleArea(int[] heights) {
        int len = heights.length;
        Deque<Integer> s = new ArrayDeque<>();
        int maxArea = 0;
        for (int i = 0; i <= len; i++) {
            int h = (i == len ? 0 : heights[i]);
            if (s.isEmpty() || h >= heights[s.peek()]) {
                s.push(i);
            } else {
                int tp = s.pop();
                int width = (s.isEmpty() ? i : i - 1 - s.peek());
                maxArea = Math.max(maxArea, heights[tp] * width);
                i--;
            }
        }
        return maxArea;
    }
}

3、复杂度分析
时间复杂度: O(m * n) 其中 dim(matrix) = (m, n)
空间复杂度: O(n) 栈

2. Solution 2

1、思路分析
思路同Solution 1，用数组模拟栈。
2、代码实现

package Q0099.Q0085MaximalRectangle;

public class Solution2 {
    /*
    执行耗时:2 ms,击败了100.00% 的Java用户
    内存消耗:41.8 MB,击败了36.24% 的Java用户
    */
    //方案2
    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[] dp = new int[n + 1];//增加一列，简化边界处理
        int res = 0;
        for (int row = 0; row < m; row++) {
            updateHeight(dp, row, matrix);

            res = Math.max(res, maxArea(dp));
        }
        return res;
    }

    public void updateHeight(int[] dp, int row, char[][] matrix) {
        for (int column = 0; column < matrix[row].length; column++) {
            dp[column] = matrix[row][column] == '1' ? dp[column] + 1 : 0;
        }
    }

    public int maxArea(int[] heights) {
        int l = 0, r = 0;
        int n = heights.length;
        int[] pos = new int[n + 1];
        pos[l] = -1;
        int res = 0;
        while (r < n) {
            while (l > 0 && heights[r] < heights[pos[l]]) {
                int h = heights[pos[l--]];
                res = Math.max(res, h * (r - pos[l] - 1));
            }
            pos[++l] = r++;
        }
        return res;
    }
}

3、复杂度分析
时间复杂度: O(m * n) 其中 dim(matrix) = (m, n)
空间复杂度: O(n) 栈

标签：Maximal,matrix,int,res,heights,length,0085,LeetCode,row
来源： https://www.cnblogs.com/junstat/p/16201379.html