I shall prove a result allows one to effortlessly construct precisely the lattices $L$ such that $height(L)=|J(L)|$.


 A closure system is a pair $(X,C)$ such that $C\subseteq P(X)$ and $C$ is closed under arbitrary intersection ($X\in C$ by taking the empty intersection). If $(X,C)$ is a closure system, then $C$ is clearly a complete lattice and elements in $L$ are said to be closed sets. It is easy to show that a pair $(X,C)$ with $C\subseteq P(X)$ is a closure system if and only if for each $R\subseteq X$ there is a smallest set $L\in C$ such that $R\subseteq L$. If $(X,C)$ is a closure system, then define $C^{*}:P(X)\rightarrow P(X)$ to be the mapping such that if $A\subseteq X$, then $C^{*}(A)$ is the smallest set in $C$ such that $A\subseteq C^{*}(A)$.

If $L$ is a finite lattice and $x\in L$, then let $J(x)$ denote the collection of all join-irreducible elements $a$ with $a\leq x$.

> $\mathbf{Theorem}$ The finite lattices such that $height(L)=|J(L)|$
> are up-to-isomorphism the finite lattices $C$ such that
> $(\{1,...,n\},C)$ is a closure system with
> $\emptyset,\{1\},...,\{1,...,n\}\in C$.

$\mathbf{Proof}$ 
Suppose that $(\{1,...,n\},C)$ is a closure system with $\emptyset,\{1\},...,\{1,...,n\}\in C$. Then $\emptyset,\{1\},...,\{1,...,n\}$ is a chain in $C$ of length $n+1$. 

Now assume that $L\in C$ is join-irreducible in the lattice $C$. Then clearly $L=\bigvee_{a\in L}^{L}C^{*}(\{a\})$. Therefore, since $L$ is join-irreducible, we have $L=C^{*}(\{a\})$ for some $a\in\{1,...,n\}$. In particular, there can be at most $n$ join-irreducible elements in the lattice $C$. We conclude that $height(L)\geq|J(L)|$, so
$height(L)=|J(L)|$, and $C^{*}(\{1\}),...,C^{*}(\{n\})$ are the join-irreducibles in $C$.

Now for the converse, assume that $L$ is a finite lattice such that $height(L)=|J(L)|$ and $J(L)=n$. Let $0=x_{0}<x_{1}<...<x_{n}=1$ be a chain of length $n+1$. Then $J(x_{0})\subset J(x_{1})\subset...\subset J(x_{n})$ (and each subset here is a proper subset). Since $J(x_{i})\subseteq J(L)$ for $0\leq i\leq n$ and $J(L)=n$, we conclude that there are $a_{1},...,a_{n}$ such that $J(x_{i})=\{a_{1},...,a_{i}\}$ for $1\leq i\leq n$. In this case, we have $J(L)=\{a_{1},...,a_{n}\}$. Now let
$C=\{J(x)|x\in L\}$. If $a\in J(L)$, then $a\in\bigcap_{i\in I}J(x_{i})$ if and only if $a\leq x_{i}$ for $i\in I$ if and only if $a\leq\bigwedge_{i\in I}x_{i}$ if and only if $a\in J(\bigwedge_{i\in I}x_{i})$. Therefore $\bigcap_{i\in I}J(x_{i})=J(\bigwedge_{i\in I}x_{i})$. We conclude that $C$ is a closure system. Furthermore, we have $J(x_{i})=\{a_{1},...,a_{i}\}\in C$ whenever $0\leq i\leq n$. I now claim that the mapping $f:L\rightarrow C$ where $f(x)=J(x)$ for all $x\in L$ is a lattice isomorphism. By definition, the mapping $L$ is surjective. Furthermore, since each $x\in L$ is a join of join-irreducible elements, we have $x=\bigvee J(x)=\bigvee f(x)$. Therefore, if $f(x)=f(y)$, then $x=\bigvee f(x)=\bigvee f(y)=y$. Therefore, the mapping $f$ is a bijection. Now, if $x\leq y$, then clearly $f(x)\subseteq f(y)$, and if $f(x)\subseteq f(y)$, then $x=\bigvee f(x)\leq\bigvee f(y)=y$. Therefore, the mapping $f$ is an order-isomorphism, and hence a lattice isomorphism. $\mathbf{QED}$