How to resolve the algorithm Reverse words in a string step by step in the V (Vlang) programming language
How to resolve the algorithm Reverse words in a string step by step in the V (Vlang) programming language
Table of Contents
Problem Statement
Reverse the order of all tokens in each of a number of strings and display the result; the order of characters within a token should not be modified.
Hey you, Bub! would be shown reversed as: Bub! you, Hey
Tokens are any non-space characters separated by spaces (formally, white-space); the visible punctuation form part of the word within which it is located and should not be modified. You may assume that there are no significant non-visible characters in the input. Multiple or superfluous spaces may be compressed into a single space. Some strings have no tokens, so an empty string (or one just containing spaces) would be the result. Display the strings in order (1st, 2nd, 3rd, ···), and one string per line. (You can consider the ten strings as ten lines, and the tokens as words.)
Let's start with the solution:
Step by Step solution about How to resolve the algorithm Reverse words in a string step by step in the V (Vlang) programming language
Source code in the v programming language
fn main() {
mut n := [
"---------- Ice and Fire ------------",
" ",
"fire, in end will world the say Some",
"ice. in say Some ",
"desire of tasted I've what From ",
"fire. favor who those with hold I ",
" ",
"... elided paragraph last ... ",
" ",
"Frost Robert -----------------------",
]
for i, s in n {
mut t := s.fields() // tokenize
// reverse
last := t.len - 1
for j, k in t[..t.len/2] {
t[j], t[last-j] = t[last-j], k
}
n[i] = t.join(" ")
}
// display result
for t in n {
println(t)
}
}
mut n := [
"---------- Ice and Fire ------------",
" ",
"fire, in end will world the say Some",
"ice. in say Some ",
"desire of tasted I've what From ",
"fire. favor who those with hold I ",
" ",
"... elided paragraph last ... ",
" ",
"Frost Robert -----------------------",
]
println(
n.map(
it.fields().reverse().join(' ').trim_space()
).join('\n')
)
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