Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area. 给定一个矩阵中,只有0和1,求出这个矩阵的一个最大的子矩阵,其中只包含1. 例如 01101 11010 01110 11110 1
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 01 0 1 1 11 1 1 1 11 0 0
问题:https://leetcode.com/problems/construct-the-rectangle/?tab=Description For a web developer, it is very important to know how to design a web page’s size. So, given a specific rectangular web page’
Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Assume that the total area
文章参考网址:http://www.iteye.com/topic/1119586 最近项目需要长方形的单个颜色选择器,google发现没有找到很多这方面的资源,比较多的是圆形的颜色选择器, 所以在参考网址上面写了一个长方形的颜色选择器。 1、首先初始化画笔,这里用到了LinearGradient线性渐变,具体代码如下: public void setInitData() {float den
题目: Given a 2D binary matrix filled with 0's and 1's, find the largest rectangle containing all ones and return its area. 题解: 在Largest Rectangle in Histogram 的神算法的基础上略作改变即可。 c++版: class Solutio
题目: Given n non-negative integers representing the histogram's bar height where the width of each bar is 1, find the area of largest rectangle in the histogram. Above is a histogram where width
Largest Rectangle in Histogram 单调栈 class Solution {public int largestRectangleArea(int[] heights) {Stack<Integer> stack=new Stack<>();int maxArea=0,n=heights.length;for(int i=0;i<=n;i++){while(!stack
84. Largest Rectangle in Histogram 暴力解法: class Solution {public int largestRectangleArea(int[] heights) {int n=heights.length;int maxArea=0;for(int i=0;i<n;i++){int min_height=heights[i];for(int j=i;
题目: A histogram is a polygon composed of a sequence of rectangles aligned at a common base line. The rectangles have equal widths but may have different heights. For example, the figure on the left s
题目:画图,学用rectangle画方形。 There is no nutrition in the blog content. After reading it, you will not only suffer from malnutrition, but also impotence. The blog content is all parallel goods. Those wh
Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. 问题关键在于重合的部分怎么计算 class So
This way 题意: 二维平面上有一些点,你现在有一个没有顶边的矩形,问你有多少种包含点的情况(每个点视为不同) 题解: 将每个点视为矩形下底边上的点,查找这个点左边有多少点,右边有多少点,这个点做完之后将其删除,相同高度的点从左到右做,对于右边的点要注意左端点位左边的点+1: 这张图就表示了相同高度右边点的可查询区间。 (刚多校结束发现2200真的是比赛中的简单题了) #incl
This way 题意: 现在有n个连着的矩形,每个矩形的宽为1,高为a[i],问你在这些矩形内部最大能组成的矩形大小。 题解: 笛卡尔树模板,模板和之前有了一些变化,增加了连边的特判,这样子就算有起始点为0的地方也无妨。当然要注意初始化 #include<bits/stdc++.h>using namespace std;#define ll long longconst i