2022-05-18:假设数组a和数组b为两组信号: 1) length(b) <= length(a); 2) 对于任意0<=i<length(b), 有b[i+1] - b[i] == a[i+1

2022-05-18:假设数组a和数组b为两组信号:

  1. length(b) <= length(a);
  2. 对于任意0<=i<length(b), 有b[i+1] - b[i] == a[i+1] - a[i]。
    那么就称信号b和信号a一致,记为b==a,
    给你好多b数组,假设有m个: b0数组、b1数组…
    给你好多a数组,假设有n个: a0数组、a1数组…
    返回一个长度为m的结果数组ans,ans[i]表示 : bi数组和多少个a数组一致。
    来自字节飞书团队。

答案2022-05-18:

前缀树。

代码用rust编写。代码如下:

use std::collections::HashMap;
use std::sync::Arc;
use std::sync::Mutex;

fn main() {
    let mut bs: Vec<Vec<isize>> = vec![vec![1, 7, 5], vec![2, 8, 6, 24]];
    let mut as0: Vec<Vec<isize>> = vec![vec![1, 7, 5], vec![2, 8, 6, 24]];
    let ans = same_teams_array(&mut bs, &mut as0);
    println!("ans = {:?}", ans);
}

fn same_teams_array(bs: &mut Vec<Vec<isize>>, as0: &mut Vec<Vec<isize>>) -> Vec<isize> {
    let m = bs.len() as isize;
    let mut root = Arc::new(Mutex::new(TrieNode::new()));
    let mut cur = Arc::clone(&mut root);
    let mut cur2 = Arc::clone(&mut cur);
    for i in 0..m {
        let k = bs[i as usize].len() as isize;
        cur = Arc::clone(&mut root);
        for j in 1..k {
            let diff = bs[i as usize][j as usize] - bs[i as usize][(j - 1) as usize];
            if !cur.lock().unwrap().nexts.contains_key(&diff) {
                cur.lock()
                    .unwrap()
                    .nexts
                    .insert(diff, Arc::new(Mutex::new(TrieNode::new())));
            }
            let c2 = Arc::clone(&mut cur.lock().unwrap().nexts.get(&diff).unwrap());
            cur = c2;
        }
        cur.lock().unwrap().indices.push(i);
    }
    let mut ans: Vec<isize> = vec![];
    for _i in 0..m {
        ans.push(0);
    }
    let n = as0.len() as isize;
    for i in 0..n {
        let k = as0[i as usize].len() as isize;
        cur = Arc::clone(&mut root);
        for j in 1..k {
            let diff = as0[i as usize][j as usize] - as0[i as usize][(j - 1) as usize];
            if !cur.lock().unwrap().nexts.contains_key(&diff) {
                break;
            }
            let c2 = Arc::clone(&mut cur.lock().unwrap().nexts.get(&diff).unwrap());
            cur = c2;
            for index in &cur.lock().unwrap().indices {
                ans[(*index) as usize] += 1;
            }
        }
    }
    return ans;
}

pub struct TrieNode {
    pub indices: Vec<isize>,
    pub nexts: HashMap<isize, Arc<Mutex<TrieNode>>>,
}

impl TrieNode {
    pub fn new() -> Self {
        Self {
            indices: vec![],
            nexts: HashMap::new(),
        }
    }
}

执行结果如下:

在这里插入图片描述


左神java代码

posted @ 2022-05-18 19:36  福大大架构师每日一题  阅读(13)  评论(0编辑  收藏  举报  来源