LeetCode——maximal-rectangle

Question

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

Solution

这个题目可以借鉴LeetCode——largest-rectangle-in-histogram 的思路，求最大矩阵的面积，所以我们只需要按行把给定’0’，‘1’矩阵，转换成对应的高度即可，然后计算最大矩阵面积。

Code

时间复杂度O(n^2).

class Solution {
public:int maximalRectangle(vector<vector<char> > &matrix) {int row = matrix.size();if (row <= 0)return 0;int col = matrix[0].size();vector<vector<int>> heights;for (int i = 0; i < row; i++) {vector<int> tmp;for (int j = 0; j < col; j++) {if (matrix[i][j] != '0') {// 累加高度if (i - 1 >= 0 && matrix[i - 1][j] != '0')matrix[i][j] += (matrix[i - 1][j] - '0');tmp.push_back(matrix[i][j] - '0');} else {// 被‘0’断开了，就需要计算一次if (tmp.size() > 0) {heights.push_back(tmp);tmp.clear();}}}if (tmp.size() > 0)heights.push_back(tmp);}int res = 0;for (int i = 0; i < heights.size(); i++) {res = max(res, calc(heights[i]));}return res;}// 给定高度，求最大矩阵面积，时间复杂度O(n)int calc(vector<int>& height) {if (height.size() <= 0)return 0;stack<int> tb;int n = height.size();int res = 0;for (int i = 0; i < n; i++) {while (!tb.empty() && height[tb.top()] >= height[i]) {int index = tb.top();tb.pop();if (tb.empty()) {res = max(res, i * height[index]);} else {res = max(res, (i - tb.top() - 1) * height[index]);}}tb.push(i);}while (!tb.empty()) {int index = tb.top();tb.pop();if (tb.empty()) {res = max(res, n * height[index]);} else {res = max(res, (n - tb.top() - 1) * height[index]);}}return res;}
};

转载于:https://www.cnblogs.com/zhonghuasong/p/7760766.html

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