原题
--- Day 1: Sonar Sweep --- You're minding your own business on a ship at sea when the overboard alarm goes off! You rush to see if you can help. Apparently, one of the Elves tripped and accidentally sent the sleigh keys flying into the ocean!
Before you know it, you're inside a submarine the Elves keep ready for situations like this. It's covered in Christmas lights (because of course it is), and it even has an experimental antenna that should be able to track the keys if you can boost its signal strength high enough; there's a little meter that indicates the antenna's signal strength by displaying 0-50 stars.
Your instincts tell you that in order to save Christmas, you'll need to get all fifty stars by December 25th.
Collect stars by solving puzzles. Two puzzles will be made available on each day in the Advent calendar; the second puzzle is unlocked when you complete the first. Each puzzle grants one star. Good luck!
As the submarine drops below the surface of the ocean, it automatically performs a sonar sweep of the nearby sea floor. On a small screen, the sonar sweep report (your puzzle input) appears: each line is a measurement of the sea floor depth as the sweep looks further and further away from the submarine.
For example, suppose you had the following report:
- 199
- 200
- 208
- 210
- 200
- 207
- 240
- 269
- 260
- 263
This report indicates that, scanning outward from the submarine, the sonar sweep found depths of 199, 200, 208, 210, and so on.
The first order of business is to figure out how quickly the depth increases, just so you know what you're dealing with - you never know if the keys will get carried into deeper water by an ocean current or a fish or something.
To do this, count the number of times a depth measurement increases from the previous measurement. (There is no measurement before the first measurement.) In the example above, the changes are as follows:
- 199 (N/A - no previous measurement)
- 200 (increased)
- 208 (increased)
- 210 (increased)
- 200 (decreased)
- 207 (increased)
- 240 (increased)
- 269 (increased)
- 260 (decreased)
- 263 (increased) In this example, there are 7 measurements that are larger than the previous measurement.
How many measurements are larger than the previous measurement?
Answer:
解析
题目要求计算深度测量值比前一次测量值增加的次数。
解答
use std::{fs::File, io::Read};
fn main() {
// 1. 读取文件
let mut file = File::open("input.txt").unwrap();
let mut content = String::new();
file.read_to_string(&mut content).unwrap();
// 2. 把文件内容转成数组
let inputs = content
.split("\n")
.map(|x| x.parse::<i32>().unwrap())
.collect::<Vec<i32>>();
// 3. 遍历数组计算后一个数比前一个数大的个数
let mut count = 0;
for i in 1..inputs.len() {
if inputs[i] > inputs[i - 1] {
count += 1;
}
}
println!("Answer: {:?}", count);
}
学习
使用 itertools 库，参考自 alexander-novo/Advent2021 (github.com)
use itertools::Itertools;
use std::fs::File;
use std::io::{self, BufRead};
fn main() {
let file = File::open("input.txt").expect("Could not read input.txt");
let depths = io::BufReader::new(file)
.lines()
// Map each line (string) to an int
.map(|line| line.unwrap().parse::<i32>().unwrap())
// tuple_windows() slides a window of two elements over the iterator of ints, grouping them into pairs.
.tuple_windows()
// Check for which pairs next (the element later on in the list) is greater
.filter(|(prev, next)| next > prev)
// The elements we have left is the number of times the list increases
.count();
println!("{}", depths)
}
