Leetcode 第85題：Maximal Rectangle--最大矩形（C++、Python）

2019.7.1  感謝張老大付費的馬拉維的小孩的視頻，張老大牛批！！！

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題目地址：Maximal Rectangle

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題目簡介：

給定一個僅包含 0 和 1 的二維二進制矩陣，找出只包含 1 的最大矩形，並返回其面積。

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

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題目解析：

按照最大正方形的方法解決問題不好解決，需要動態的更新寬和高。根據discuss區，看到使用Largest Rectangle in Histogram方法。即將matrix的沒一層看作底，計算最大直方圖。（真不知道如果不是剛做84怎麼想起來的）

C++：

class Solution {
public:
    int maximalRectangle(vector<vector<char>>& matrix) {
        if (matrix.size() == 0 || matrix[0].size() == 0)
            return 0;
        int _size = matrix[0].size();
        vector<int> height(_size + 1, 0);
        int ans = 0;
        for (int i = 0; i < matrix.size(); i++)
        {
            for (int j = 0; j < _size; j++)
                height[j] = matrix[i][j] == '1' ? height[j] + 1 : 0;
            stack<int> st;
            st.push(-1);
            for (int k = 0; k < _size + 1; k++)
            {
                while((st.top() != -1 && height[k] < height[st.top()]) || (height[k] < height[_size]))
                {
                    int h = height[st.top()];
                    st.pop();
                    int w = k - 1 - st.top();
                    ans = max(ans, h * w);
                }
                st.push(k);
            }
        }
        return ans;
    }
};

Python：

class Solution:
    def maximalRectangle(self, matrix):
        if not matrix or not matrix[0]:
            return 0
        n = len(matrix[0])
        height = [0] * (n + 1)
        ans = 0
        for row in matrix:
            for i in range(n):
                height[i] = height[i] + 1 if row[i] == '1' else 0
            stack = [-1]
            for i in range(n + 1):
                while height[i] < height[stack[-1]]:
                    h = height[stack.pop()]
                    w = i - 1 - stack[-1]
                    ans = max(ans, h * w)
                stack.append(i)
        return ans