答案1
欢迎!基于 Zarko 的最近的答案:
\documentclass[tikz, margin=3mm]{standalone}
\usetikzlibrary{chains, fit, positioning,calc}
\begin{document}
% from https://tex.stackexchange.com/a/409901/121799
\begin{tikzpicture}[
node distance = 0pt,
start chain = A going right,
start chain = B going right,
base/.style = {draw,rectangle,minimum height=7mm, outer sep=0pt,fill=green},
boxA/.style = {base,minimum width=15mm, on chain=A,fill=green},
boxB/.style = {base,minimum width=15mm, on chain=B,fill=green}
]
\node[base, on chain=A] (nodeAa) {$2/5/8$};
\node[boxA] (nodeAb) {$1/7/9$};
\node[boxA] (nodeAc) {$2/3/9$};
\node[boxA] (nodeAd) {$2/7/12$};
\node[boxA] (nodeAe) {$1/15/15$};
\node[boxA] (nodeAf) {$2/9/15$};
\node[boxA] (nodeAg) {$1/8/17$};
%
\node[thick,inner sep=0pt, fit=(A-1) (A-7)] {};
%
\node[base, on chain=B,below=1cm of nodeAa] (nodeBa) {$2/5/8$};
\node[boxB] (nodeBb) {$1/7/9$};
\node[boxB] (nodeBc) {$2/3/9$};
\node[boxB] (nodeBd) {$2/7/12$};
\node[boxB] (nodeBe) {$1/15/15$};
\node[boxB] (nodeBf) {$2/9/15$};
\node[boxB] (nodeBg) {$1/8/17$};
%
\node[thick,inner sep=0pt, fit=(B-1) (B-7)] {};
%
\draw[-latex] ($(nodeAa.south)-(0,0.7cm)$) -- (nodeAa) node[midway,left]{First Element};
\draw[-latex] ($(nodeAd.south east)-(0.2cm,0.7cm)$) -- ($(nodeAd.south east)-(0.2cm,0)$)
node[midway,right]{Second Element};
\draw[latex-latex,out=30,in=150] (nodeAd.north) to (nodeAe.north east);
\draw[latex-latex,out=150,in=30] (nodeAd.north) to (nodeAc.north west);
\draw[-latex] ($(nodeBa.south)-(0,0.7cm)$) -- (nodeBa) node[midway,right]{OK Element};
\draw[-latex] (nodeAd.south) to (nodeBc.north);
\draw[-latex] (nodeAc.south) to (nodeBd.north);
\end{tikzpicture}
\end{document}