2022-12-18：给定一个长度为n的二维数组graph，代表一张图， graph[i] = {a,b,c,d} 表示i讨厌(a,b,c,d)，讨厌关系为双向的， 一共有n个人，编号0~n-1， 讨

Posted 2022-12-19 福大大架构师每日一题

2022-12-18：给定一个长度为n的二维数组graph，代表一张图，
graph[i] = a,b,c,d 表示i讨厌(a,b,c,d)，讨厌关系为双向的，
一共有n个人，编号0~n-1，
讨厌的人不能一起开会。
返回所有人能不能分成两组开会。
来自微软面试。

答案2022-12-18：

力扣785。并查集。

代码用rust编写。代码如下：

use rand::Rng;
use std::iter::repeat;
fn main() 
    let mut graph: Vec<Vec<i32>> = vec![vec![1, 2, 3], vec![0, 2], vec![0, 1, 3], vec![0, 2]];
    let ans = is_bipartite(&mut graph);
    println!("ans = ", ans);


fn is_bipartite(graph: &mut Vec<Vec<i32>>) -> bool 
    let n = graph.len() as i32;
    let mut uf = UnionFind::new(n);
    for neighbours in graph.iter() 
        for i in 1..neighbours.len() as i32 
            uf.union(neighbours[(i - 1) as usize], neighbours[i as usize]);
        
    
    for i in 0..n 
        for j in graph[i as usize].iter() 
            if uf.same(i, *j) 
                return false;
            
        
    
    return true;


struct UnionFind 
    f: Vec<i32>,
    s: Vec<i32>,
    h: Vec<i32>,

impl UnionFind 
    pub fn new(n: i32) -> Self 
        let mut f: Vec<i32> = repeat(0).take(n as usize).collect();
        let mut s: Vec<i32> = repeat(0).take(n as usize).collect();
        let mut h: Vec<i32> = repeat(0).take(n as usize).collect();
        for i in 0..n 
            f[i as usize] = i;
            s[i as usize] = 1;
        
        Self  f, s, h 
    

    fn find(&mut self, mut i: i32) -> i32 
        let mut hi = 0;
        while i != self.f[i as usize] 
            self.h[hi as usize] = i;
            hi += 1;
            i = self.f[i as usize];
        
        while hi > 0 
            hi -= 1;
            self.f[self.h[hi as usize] as usize] = i;
        
        return i;
    

    pub fn same(&mut self, i: i32, j: i32) -> bool 
        return self.find(i) == self.find(j);
    

    pub fn union(&mut self, i: i32, j: i32) 
        let mut fi = self.find(i);
        let mut fj = self.find(j);
        if fi != fj 
            if self.s[fi as usize] >= self.s[fj as usize] 
                self.f[fj as usize] = fi;
                self.s[fi as usize] = self.s[fi as usize] + self.s[fj as usize];
             else 
                self.f[fi as usize] = fj;
                self.s[fj as usize] = self.s[fi as usize] + self.s[fj as usize];
            
        
    

执行结果如下：

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左神java代码