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Some fixes

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Antti H S Laaksonen 2017-02-27 21:29:32 +02:00
parent 074134ac54
commit 98fda0b259
8 changed files with 54 additions and 48 deletions
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12
chapter03.tex
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@ -664,7 +664,7 @@ string s = "monkey";
sort(s.begin(), s.end());
\end{lstlisting}
Sorting a string means that the characters
in the string are sorted.
of the string are sorted.
For example, the string ''monkey'' becomes ''ekmnoy''.
\subsubsection{Comparison operators}
@ -775,7 +775,7 @@ an element $x$ in the array \texttt{t}:
\begin{lstlisting}
for (int i = 1; i <= n; i++) {
if (t[i] == x) // x found at index i
if (t[i] == x) {} // x found at index i
}
\end{lstlisting}
@ -816,7 +816,7 @@ The above idea can be implemented as follows:
int a = 1, b = n;
while (a <= b) {
int k = (a+b)/2;
if (t[k] == x) // x found at index k
if (t[k] == x) {} // x found at index k
if (t[k] > x) b = k-1;
else a = k+1;
}
@ -850,7 +850,7 @@ int k = 1;
for (int b = n/2; b >= 1; b /= 2) {
while (k+b <= n && t[k+b] <= x) k += b;
}
if (t[k] == x) // x was found at index k
if (t[k] == x) {} // x was found at index k
\end{lstlisting}
The variables $k$ and $b$ contain the position
@ -893,7 +893,7 @@ $\texttt{ok}(x)$ & \texttt{false} & \texttt{false}
\end{center}
\noindent
Now, the value $k$ can be found using binary search:
Now, the value of $k$ can be found using binary search:
\begin{lstlisting}
int x = -1;
@ -923,7 +923,7 @@ the total time complexity is $O(n \log z)$.
Binary search can also be used to find
the maximum value for a function that is
first increasing and then decreasing.
Our task is to find a value $k$ such that
Our task is to find a position $k$ such that
\begin{itemize}
\item