【leetcode】85. Maximal Rectangle 0/1矩阵的最大全1子矩阵

1. 题目

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

For example, given the following matrix:

1 0 1 0 0
1 0 1 1 1
1 1 1 1 1
1 0 0 1 0
Return 6.

2. 思路

将m*n的矩阵，转换成为m个数组，既可以用上一题的方法求解。【leetcode】84. Largest Rectangle in Histogram 最大面积的覆盖矩阵

参考了http://blog.csdn.net/doc_sgl/...

3. 代码

耗时：19ms

class Solution {
public:
    // 对m*n的矩阵，转化为m个84题的一维数组的最大矩形问题。
    // 转化方法是，对当前行转换为当前位置的上方连续1的个数作为一个高度值
    int maximalRectangle(vector<vector<char>>& matrix) {
        vector<vector<int>> matrix2;
        int m = matrix.size();
        if (m == 0) { return 0; }
        int n = matrix[0].size();
        if (n == 0) {return 0; }
        for (int i = 0; i < m; i++) {
            vector<int> row;
            for (int j = 0; j < n; j++) {
                row.push_back(matrix[i][j] == '0' ? 0 : 1);
            }
            matrix2.push_back(row);
        }
        return maximalRectangle(matrix2);
    }
    int maximalRectangle(vector<vector<int>>& matrix) {
        int m = matrix.size();
        if (m == 0) { return 0; }
        int n = matrix[0].size();
        if (n == 0) {return 0; }
        int max_area = largestRectangleArea(matrix[0]);
        for (int i = 1; i < m; i++) {
            vector<int>& line_up = matrix[i-1];
            vector<int>& line = matrix[i];
            for (int j = 0; j < n; j++) {
                if (line[j] == 1) { line[j] = line_up[j] + 1; }
            }
            int area = largestRectangleArea(line);
            //cout << "i=" << i << " area=" << area << endl;
            if (area > max_area) { max_area = area; }
        }
        return max_area;
    }
    
    int largestRectangleArea(vector<int>& heights) {
        int max_area = 0;
        heights.push_back(0);
        int sz = heights.size();
        int stack[sz];
        stack[0] = heights[0];
        int stack_idx = 0;
        int i = 1;
        while (stack_idx >= 0 && i < sz) {
            if (heights[i] >= stack[stack_idx]) {
                stack[++stack_idx] = heights[i++];
                continue;
            }
            while (stack_idx >= 0 && stack[stack_idx] > heights[i]) {
                int area = stack[stack_idx] * (i - stack_idx);
                if (area > max_area) { max_area = area; }
                stack_idx--;
            }
            while (stack_idx < i) {
                stack[++stack_idx] = heights[i];
            }
            i++;
        }
        return max_area;
    }
};