LeetCode: Maximal Rectangle

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.

地址：https://oj.leetcode.com/problems/maximal-rectangle/

算法：要解决这道题，得利用Largest Rectangle in Histogram这道题的解法。我们把矩阵的每一行都看成histogram's bar的底，比如第i行可以看成从最后一行到第i行之间的histogram，并且每一列的高度为在该列上从i行开始向下数连续为1的行数。这样以第i行为矩阵的上面一条边的最大矩阵的值即为这个histogram中最大的矩阵面积。也就是说对于没一个行，我们都可以计算出该行所代表的histogram，然后调用Largest Rectangle in Histogram里面的函数就可以找出以该行为上边的最大矩阵，然后从所有行中找到整个问题的最大矩阵。代码：

class Solution {
public:
    int maximalRectangle(vector<vector<char> > &matrix) {
        if(matrix.empty() || matrix[0].empty()) return 0;
        int row_num = matrix.size();
        int col_num = matrix[0].size();
        vector<int> histogram1(col_num), histogram2(col_num);
        for(int j = 0; j < col_num; ++j){
            histogram1[j] = matrix[row_num-1][j] - '0';
        }
        int max_area = largestRectangleArea(histogram1);
        vector<int> *p_now = &histogram2;
        vector<int> *p_pre = &histogram1;
        for(int i = row_num-2; i >= 0; --i){
            for(int j = 0; j < col_num; ++j){
                if(matrix[i][j] == '0'){
                    (*p_now)[j] = 0;
                }else if(matrix[i+1][j] == matrix[i][j]){
                    (*p_now)[j] = (*p_pre)[j] + 1;
                }else{
                    (*p_now)[j] = 1;
                }
            }
            int area = largestRectangleArea(*p_now);
            if(area > max_area)
                max_area = area;
            vector<int> *temp = p_now;
            p_now = p_pre;
            p_pre = temp;
        }
        return max_area;
    }
    int largestRectangleArea(vector<int> &height) {
        int len = height.size();
        if(len < 1) return 0;
        stack<int> stk;
        int i = 0;
        int max_area = 0;
        while(i < len){
            if(stk.empty() || height[stk.top()] <= height[i]){
                stk.push(i++);
            }else{
                int t = stk.top();
                stk.pop();
                int area = height[t] * (stk.empty() ? i : i - stk.top() - 1);
                if(area > max_area){
                    max_area = area;
                }
            }
        }
        while(!stk.empty()){
            int t = stk.top();
            stk.pop();
            int area = height[t] * (stk.empty() ? len : len - stk.top() - 1);
            if(area > max_area){
                max_area = area;
            }
        }
        return max_area;
    }
};