Commit 5c0fc8c

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Fix lints

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+10

-8

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+10

-8

lines changed

Lines changed: 1 addition & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -23,7 +23,7 @@ pub fn parse(input: &str) -> Vec<Room<'_>> {`
`23`	`23`	`let size = line.len();`
`24`	`24`	`let name = &line[..size - 11];`
`25`	`25`	`let sector_id = (&line[size - 10..size - 7]).unsigned();`
`26`		`- let checksum = line[size - 6..size - 1].as_bytes();`
	`26`	`+ let checksum = &line.as_bytes()[size - 6..size - 1];`
`27`	`27`
`28`	`28`	// Count the frequency of each digit, the frequency of each frequency `fof` and the
`29`	`29`	`// highest total frequency.`

Lines changed: 2 additions & 2 deletions

Original file line number	Diff line number	Diff line change
`@@ -6,7 +6,7 @@`
`6`	`6`	`//! The trick for part two is that the plants will eventually stabilize into a repeating pattern`
`7`	`7`	`//! that expands by the same amount each generation. Once 2 deltas between 3 subsequent`
`8`	`8`	`//! generations are the same we extrapolate 50 billion generations into the future.`
`9`		`-use std::iter::repeat;`
	`9`	`+use std::iter::repeat_n;`
`10`	`10`
`11`	`11`	`pub struct Input {`
`12`	`12`	`rules: Vec<usize>,`
`@@ -73,7 +73,7 @@ fn step(rules: &[usize], tunnel: &Tunnel) -> Tunnel {`
`73`	`73`	`let mut plants = Vec::with_capacity(1_000);`
`74`	`74`
`75`	`75`	`// Add four extra empty pots to the end to make checking the last pattern easier.`
`76`		`- for plant in tunnel.plants.iter().chain(repeat(&0).take(4)) {`
	`76`	`+ for plant in tunnel.plants.iter().copied().chain(repeat_n(0,4)) {`
`77`	`77`	`index = ((index << 1) \| plant) & 0b11111;`
`78`	`78`
`79`	`79`	`sum += position * rules[index] as i64;`

Lines changed: 5 additions & 3 deletions

Original file line number	Diff line number	Diff line change
`@@ -23,9 +23,11 @@ type Input = [Axis; 3];`
`23`	`23`	`/// Group each axis together`
`24`	`24`	`pub fn parse(input: &str) -> Input {`
`25`	`25`	`let n: Vec<_> = input.iter_signed().collect();`
`26`		`- [[n[0], n[3], n[6], n[9], 0, 0, 0, 0], [n[1], n[4], n[7], n[10], 0, 0, 0, 0], [`
`27`		`- n[2], n[5], n[8], n[11], 0, 0, 0, 0,`
`28`		`- ]]`
	`26`	`+ [`
	`27`	`+ [n[0], n[3], n[6], n[9], 0, 0, 0, 0],`
	`28`	`+ [n[1], n[4], n[7], n[10], 0, 0, 0, 0],`
	`29`	`+ [n[2], n[5], n[8], n[11], 0, 0, 0, 0],`
	`30`	`+ ]`
`29`	`31`	`}`
`30`	`32`
`31`	`33`	`pub fn part1(input: &Input) -> i32 {`

Lines changed: 1 addition & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -64,7 +64,7 @@ pub struct Input {`
`64`	`64`	`/// State of a single exploration path through the valves.`
`65`	`65`	`///`
`66`	`66`	/// * `todo` Binary mask of unopened valves. For example if there are 3 unopened valves left this
`67`		-/// could look like `11100`.
	`67`	+/// could look like `11100`.
`68`	`68`	/// * `from` Index of current valve.
`69`	`69`	/// * `time` The remaining time left.
`70`	`70`	/// * `pressure` Total pressure released from all opened valves including future extrapolated flow.

Lines changed: 1 addition & 1 deletion

Original file line number	Diff line number	Diff line change
`@@ -56,7 +56,7 @@`
`56`	`56`	//! * `B` and `F` are both supported only by `A` so their depth is 1.
`57`	`57`	//! * `C` will fall if either `A` or `B` is removed so its depth is 2.
`58`	`58`	//! * `D` will only fall when `A` is removed. Removing `F` would leave it supported by `B` and `C`
`59`		-//! or vice-versa. The common ancestor of the path to the root is `A` so its depth is 1.
	`59`	+//! or vice-versa. The common ancestor of the path to the root is `A` so its depth is 1.
`60`	`60`	//! * `E`'s common ancestor is the floor so its depth is 0.
`61`	`61`	`//!`
`62`	`62`	//! In total `0 (A) + 0 (G) + 1 (B) + 1 (F) + 2 (C) + 1 (D) + 0 (E) = 5` bricks will fall.

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