如何绘制带箭头的数组？

Question

我想画这个：

到目前为止，我可以画这么多，但红色箭头却很难画

\documentclass{article} 
\usepackage{etoolbox}
\begin{document}
\begin{center}
\begin{tabular}{ |c|c|c|c|c|c|c|c| } 
\hline
 3 & 4 & 2 & 1 & 4 & 1 & 2 & 3  \\
\hline
\hline 
\end{tabular}
\end{center}
\end{document}

Answer 1

如果你接受纯TikZ解决方案：

\documentclass[tikz,border=2mm]{standalone} 
\usetikzlibrary{positioning,matrix, arrows.meta}

\begin{document}
\begin{tikzpicture}
\matrix (A) [matrix of nodes, nodes={draw, minimum size=8mm},
    column sep=-\pgflinewidth]{
    3 & 4 & 2 & 1 & 4 & 1 & 2 & 3\\};
\foreach \i [evaluate=\i as \ni using {int(\i)},
                evaluate=\i as \ntext using {int(\i-1)}] in {1,3,4,6} 
    \draw [{Stealth}-, red!70] (A-1-\ni.south west)--++(-90:5mm) node[below] {\ntext};
\end{tikzpicture}
\end{document}

Answer 2

使用{NiceTabular}和nicematrixTikZ 创建的 PGF/TikZ 节点绘制箭头nicematrix。

\documentclass{article}
\usepackage{nicematrix,tikz}
\usetikzlibrary{arrows.meta}

\begin{document}

\begin{NiceTabular}{*{8}{c}}[hvlines]
  3 & 4 & 2 & 1 & 4 & 1 & 2 & 3 
\CodeAfter
  \begin{tikzpicture} [{Stealth}-,red]
    \draw (2-|1) -- ++(0,-4mm) node [below] {0} ; 
    \draw (2-|3) -- ++(0,-4mm) node [below] {2} ; 
    \draw (2-|4) -- ++(0,-4mm) node [below] {3} ; 
    \draw (2-|6) -- ++(0,-4mm) node [below] {5} ; 
  \end{tikzpicture}
\end{NiceTabular}

\end{document}

Answer 3

还有一个普通的 TikZ 解决方案：

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}[thick,scale=.6]
\draw (0,0) grid (8,1); 
\path (.5,.5) node{$3$} ++(1,0) node{$4$} 
++(1,0) node{$2$} ++(1,0) node{$1$} ++(1,0) node{$4$} 
++(1,0) node{$1$} ++(1,0) node{$2$} ++(1,0) node{$3$};
\foreach \i in {0,2,3,5}
\draw[latex-,red] (\i,0)--+(-90:1) node[below,black]{$\i$};
\end{tikzpicture}
\end{document}

PS：您可以使用foreach来path缩短代码

\path (.5,.5) node{$3$} foreach \i in {4,2,1,4,1,2,3} {++(1,0) node{$\i$}};