Make changes suggested by Clippy.

source/add_binary/mod.rs

@@ -9,9 +9,8 @@ pub fn add_binary(a: String, b: String) -> String {
   let mut sum = String::new();
 
   // Create reversed iterators for both binary strings.
-  let (mut a, mut b) = (a.chars().rev(), b.chars().rev());
+  let mut b = b.chars().rev();
-
+  for a in a.chars().rev() {
-  while let Some(a) = a.next() {
     // Safe to unwrap since we know both iterators will be the same length.
     let b = b.next().unwrap();
 

source/longest_common_prefix/mod.rs

@@ -1,7 +1,6 @@
-pub fn longest_common_prefix(strings: Vec<String>) -> String {
+pub fn longest_common_prefix(mut strings: Vec<String>) -> String {
   // Sort the strings ascending by length.
-  let mut strings = strings.clone();
+  strings.sort_by_key(|a| a.len());
-  strings.sort_by(|a, b| a.len().cmp(&b.len()));
 
   // Then grab the shortest string.
   let shortest_string = match strings.first() {

source/plus_one/mod.rs

@@ -8,7 +8,7 @@ pub fn plus_one(digits: Vec<i32>) -> Vec<i32> {
   // Create a vector to store the incremented digits.
   let mut incremented = vec![];
 
-  while let Some(digit) = digits.next() {
+  for digit in digits.by_ref() {
     if digit == 9 {
       // When the current digit is 9 save 0 and keep going through the digits.
       incremented.push(0);

source/valid_anagram/mod.rs

@@ -6,10 +6,10 @@ pub fn is_anagram(a: String, b: String) -> bool {
 
   // Sort both strings' characters so they're in alphabetical order.
   let mut a = a.chars().collect::<Vec<_>>();
-  a.sort();
+  a.sort_unstable();
 
   let mut b = b.chars().collect::<Vec<_>>();
-  b.sort();
+  b.sort_unstable();
 
   // And if they are equal, then they're anagrams.
   a == b

source/valid_parenthesis/mod.rs

@@ -35,5 +35,5 @@ pub fn is_valid(string: String) -> bool {
 
   // After the string has been looped over, if anything is left in the stack the
   // string is invalid.
-  stack.len() == 0
+  stack.is_empty()
 }