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I have two (2) questions. QUESTION 1 I have these code where the table that includes: Fetch value from database. Table with MultiCells. Text fits inside table cell. I create three (3) table, data from database and the result shows like table below. Code seem to break into new line after the first table. My goal is to make the table look like this Here's my current code Code.php class myPDF extends FPDF { // CALCULATE var $widths; var $aligns; function SetWidths($w) { //Set the array of column widths $this->widths=$w; } function SetAligns($a) { //Set the array of column alignments $this->aligns

1.场景还原 近期项目中的excel导入导出功能需求频繁的出现，趁此机会，今天笔者对POI的Excel数据的导入导出做一番详解，希望对大家有所帮助。 2.准备工作 ①添加POI依赖 <!-- Excel 操作包 --> <dependency> <groupId> org.apache.poi </groupId> <artifactId> poi-ooxml </artifactId> <version> 3.14-beta1 </version> </dependency> <dependency> <groupId> org.apache.poi </groupId> <artifactId> poi-ooxml-schemas </artifactId> <version> 3.14-beta1 </version> </dependency> <dependency> <groupId> org.apache.poi </groupId> <artifactId> poi </artifactId> <version> 3.14-beta1 </version> </dependency> 以及excel Jar包依赖 <dependency> <groupId> net.sourceforge.jexcelapi </groupId> <artifactId> jxl

There is a ball in a maze with empty spaces and walls. The ball can go through empty spaces by rolling up, down, left or right, but it won't stop rolling until hitting a wall. When the ball stops, it could choose the next direction. Given the ball's start position, the destination and the maze, find the shortest distance for the ball to stop at the destination. The distance is defined by the number of empty spaces traveled by the ball from the start position (excluded) to the destination (included). If the ball cannot stop at the destination, return -1. The maze is represented by a binary 2D

https://leetcode.com/problems/01-matrix/#/description 给一个二维01数组，找出每个位置和最近的0的距离 BFS 先把所有0入队，把1置为MAX_VALUE，然后把最靠近0的1的距离算出来，然后将他们入队，再算距离最靠近0的1的1的距离算出来，依次处理 public class Solution { public List<List<Integer>> updateMatrix(List<List<Integer>> matrix) { int m = matrix.size(); int n = matrix.get(0).size(); Queue<int[]> queue = new LinkedList<>(); for (int i = 0; i < m; i++) { for (int j = 0; j < n; j++) { if (matrix.get(i).get(j) == 0) { queue.offer(new int[] {i, j}); } else { matrix.get(i).set(j, Integer.MAX_VALUE); } } } int[][] dirs = {{-1, 0}, {1, 0}, {0, -1}, {0, 1}}; while (!queue.isEmpty()) {