题目地址:https://leetcode.com/problems/maximal-rectangle/description/ 题目描述: Given a 2D binary matrix filled with 0’s and 1’s, find the largest rectangle containing only 1’s and return its area. Example: Input:[
代码: 1classSolution {2public:3intmaximalRectangle(vector<vector<char> > &matrix) {4if(matrix.empty() || matrix[0].empty())return0;5introw_num =matrix.size();6intcol_num = matrix[0].size();7vector<int>histogram1(col_num), histogram2(col_num);8for(intj =0; j < col_num; ++...
计算出最大面积,这个过程可以借用栈来完成; 1classSolution {2public:3intmaximalRectangle(vector<vector<char> > &matrix) {4//Start typing your C/C++ solution below5//DO NOT write int main() function6intres =0;7if(matrix.empty() || matrix[0].empty())8returnres;9intm =matrix.size();10...
publicclassSolution {publicintmaximalRectangle(char[][] matrix) {if(matrix.length<=0)return0;introw =matrix.length;intcol = matrix[0].length;int[] left =newint[col];int[] right =newint[col];int[] height =newint[col];intres = 0;//将 right 数组初始化为 列的宽度for(inti=0;i<col...
Maximal Rectangle [leetcode] 的三种思路 第一种方法是利用DP。时间复杂度是 O(m * m * n) dp(i,j):矩阵中同一行以(i,j)结尾的所有为1的最长子串长度 代码例如以下: int maximalRectangle(vector<vector<char> > &matrix) { int m = matrix.size();...
LeetCode: Maximal Rectangle 这道题把二维图转换成行数个的柱状图题,就简单了 1classSolution {2public:3intmaximalRectangle(vector<vector<char> > &matrix) {4intx =matrix.size();5if(0== x)return0;6inty = matrix[0].size();7if(0== y)return0;8vector<vector<int> > result(x, vector<...
[LeetCode] 85. Maximal Rectangle 最大矩形 铭信 MtF/想入CS/偷偷关注知乎所有小姐姐 6 人赞同了该文章 题目:二维矩阵由0或1组成,求仅包含1的矩形最大面积 很明显,图中灰色矩形面积最大,答案为6。 这道题可以用动态规划一行一行来处理,但是网上资料我看了很久才明白。所以讲一下自己的理解,希望更通俗易懂...
**/publicclassMaximalRectangle {privateintlargestRectangleArea(int[] height){int[] min =newint[height.length];//记录i~j的最小高度intmax = 0;for(inti = 0; i < height.length; i++) {//最好的情况都无法超过最大面积,则跳过if(height[i] > 0 && max/height[i] >= height.length - i)...
1 <= row, cols <= 200 matrix[i][j] 为'0' 或'1' 题目难度:困难 通过次数:216.3K 提交次数:388K 贡献者:LeetCode 相关标签 相似题目 C++ 1 class Solution { 2 public: 3 int maximalRectangle(vector<vector<char>>& matrix) { 4 5 } 6 }; 您必须登录后才能提交解答! 提交解答©...
1publicclassSolution {2publicintlargestRectangleArea(int[] height) {3if(height ==null|| height.length == 0)4return0;5Stack<Integer> index =newStack<Integer>();6inttotalMax = 0;7ArrayList<Integer> newHeight =newArrayList<Integer>();8for(inti:height) newHeight.add(i);9newHeight.add(0)...