Leetcode | Maximal Rectangle

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

穷举很难。之前做Largest Rectangle in Histogram的时候，看到网上有人说这两道题很类似。于是开始往这方面想。

思路如下：

1. 对于第j列，每一行可以算出从j这个位置开始，连续的'1'的个数。这样在第j列就可以计算得到一个histogram。

2. 然后直接应用Largest Rectangle in Histogram的O(n)算法求最大值；

3. 所有最大值再求最大，就是结果。

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int n = height.size();
        if (n == 0) return 0;
        
        int max = 0, last, area;
        
        stack<int> indexes;
        height.push_back(0);
        
        for (int i = 0; i < height.size(); ) {
            if (indexes.empty() || (height[i]>=height[indexes.top()])) {
                indexes.push(i);
                i++;
            } else {
                last = height[indexes.top()];
                indexes.pop();
                area = last * (indexes.empty() ? i : i - indexes.top() - 1);
                if (area > max) max = area;
            }
        }
        
        return max;
    }
    int maximalRectangle(vector<vector<char> > &matrix) {
        int m = matrix.size();
        if (m == 0) return 0;
        int n = matrix[0].size();
        if (n == 0) return 0;
        
        vector<int> hist(m, 0);
        int max = 0, area;
        
        for (int j = 0; j < n; ++j) {
            for (int i = 0; i < m; ++i) {
                hist[i] = 0; 
                for (int k = j; k < n && matrix[i][k] == '1'; ++k) hist[i]++;  
            }        
            area = largestRectangleArea(hist);
            if (area > max) max = area;
        }
        return max;
    }
};

当然这样建histogram的复杂度是O(n*m*n)。

建historgram可以事先建好。动态规划啊，竟然没想到。。。。

class Solution {
public:
    int largestRectangleArea(vector<int> &height) {
        int n = height.size();
        if (n == 0) return 0;
        
        int max = 0, last, area;
        
        stack<int> indexes;
        height.push_back(0);
        
        for (int i = 0; i < height.size(); ) {
            if (indexes.empty() || (height[i]>=height[indexes.top()])) {
                indexes.push(i);
                i++;
            } else {
                last = height[indexes.top()];
                indexes.pop();
                area = last * (indexes.empty() ? i : i - indexes.top() - 1);
                if (area > max) max = area;
            }
        }
        
        return max;
    }
    int maximalRectangle(vector<vector<char> > &matrix) {
        int m = matrix.size();
        if (m == 0) return 0;
        int n = matrix[0].size();
        if (n == 0) return 0;
        
        vector<vector<int> > hists(n + 1, vector<int>(m, 0));
       
        for (int j = n - 1; j >= 0; --j) {
            for (int i = 0; i < m; ++i) {
                if (matrix[i][j] == '0') {
                    hists[j][i] = 0;
                } else {
                    hists[j][i] = hists[j + 1][i] + 1;
                }
            }        
        }
        
        int max = 0, area;    
        for (int j = n - 1; j >= 0; --j) {
            area = largestRectangleArea(hists[j]);
            if (area > max) max = area;
        }
        
        return max;
    }
};

 这样就是O(m*n)了。空间复杂度是O(m*n)。