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;}
};