Newbee (#3)

1. 新手村题目
2. 调整一下macro
3. 并查集

Author: hxzhao527

datastructure/src/list.rs

@@ -1,5 +1,24 @@
 #[derive(PartialEq, Eq, Clone, Debug)]
 pub struct ListNode {
-  pub val: i32,
-  pub next: Option<Box<ListNode>>
-}
+    pub val: i32,
+    pub next: Option<Box<ListNode>>,
+}
+
+// impl TryFrom<&[i32]> for ListNode {
+//   type Error = &'static str;
+
+//   fn try_from(s: &[i32]) -> Result<Self, Self::Error> {
+        
+//         if s.is_empty() {
+//             return Err("empty");
+//         }
+
+//         let head = ListNode {
+//             val: s.first().copied().unwrap(),
+//             next: Self::try_from(&s[1..]).ok(),
+//         };
+//         Ok(head)
+//   }
+  
+// }
+

macros/src/lib.rs

@@ -8,14 +8,14 @@ use proc_macro::TokenStream;
 mod tree_impl;
 mod list_impl;
 
-/// tree!(tree-define)
+/// tree!(tree-define) => Option<Rc<RefCell<TreeNode>>>
 /// tree-define: {val: int, [left: tree-define], [right: tree-define] }
 #[proc_macro]
 pub fn tree(input: TokenStream) -> TokenStream {
     tree_impl::tree(input.into()).into()
 }
 
-/// list!(1,2,3,4)
+/// list!(1,2,3,4) => Option<Box<ListNode>>
 /// list!([1,2,3,4])
 #[proc_macro]
 pub fn list(input: TokenStream) -> TokenStream{

macros/src/tree_impl.rs

@@ -3,8 +3,13 @@ use std::str::FromStr;
 
 
 pub fn tree(input: TokenStream) -> TokenStream {
-    let tree = TreeNode::from(input);
-    TokenStream::from_str(format!("{}", tree).as_str()).unwrap()
+    if input.is_empty(){
+        TokenStream::from_str("None").unwrap()
+    } else {
+        let tree = TreeNode::from(input);
+        TokenStream::from_str(format!("Some(::std::rc::Rc::new(::std::cell::RefCell::new({})))", tree).as_str()).unwrap()
+    }
+    
 }
 
 #[derive(Debug)]

src/array/ext.rs

@@ -14,3 +14,5 @@ pub mod rng;
 // pub mod topological;
 
 pub mod trie;
+
+pub mod disjoint;

src/array/ext/disjoint.rs

@@ -0,0 +1,152 @@
+//! # Disjoint-set data structure
+//!
+//! wiki: [Disjoint-set data structure](https://en.wikipedia.org/wiki/Disjoint-set_data_structure)
+//!
+//! 效果: 将数据分组
+//!
+//! ## 题目
+//! * 简单
+//! * 中等
+//!     * [547. 省份数量](find_circle_num)
+//!
+
+struct UnionFind {
+    count: usize,                           // 连通分量的个数, 总共分成了几组
+    parent: std::cell::RefCell<Vec<usize>>, // 记录每个节点的父节点，父节点为自身的是根节点
+    size: Vec<usize>,                       // 记录每个连通分量的大小
+}
+
+impl UnionFind {
+    pub fn new(size: usize) -> Self {
+        Self {
+            count: size,
+            parent: std::cell::RefCell::new((0..size).collect()),
+            size: vec![1; size],
+        }
+    }
+    pub fn count(&self) -> usize {
+        self.count
+    }
+    pub fn find(&self, p: usize) -> usize {
+        let mut root = p;
+        while root != self.parent.borrow()[root] {
+            root = self.parent.borrow()[root];
+        }
+        let mut p = p;
+        while p != root {
+            let next = self.parent.borrow()[p];
+            self.parent.borrow_mut()[p] = root;
+            p = next;
+        }
+        root
+    }
+    pub fn is_connected(&self, p: usize, q: usize) -> bool {
+        self.find(p) == self.find(q)
+    }
+    pub fn connect(&mut self, p: usize, q: usize) {
+        let (p_root, q_root) = (self.find(p), self.find(q));
+        if p_root == q_root {
+            return;
+        }
+        if self.size[p_root] < self.size[q_root] {
+            self.parent.borrow_mut()[p_root] = q_root;
+            self.size[q_root] += self.size[p_root];
+        } else {
+            self.parent.borrow_mut()[q_root] = p_root;
+            self.size[p_root] += self.size[q_root];
+        }
+        self.count -= 1;
+    }
+}
+
+/// [547. 省份数量](https://leetcode.cn/problems/number-of-provinces/)
+pub fn find_circle_num(is_connected: Vec<Vec<i32>>) -> i32 {
+    let mut uf = UnionFind::new(is_connected.len());
+    is_connected.into_iter().enumerate().for_each(|(i, line)| {
+        line.into_iter().enumerate().for_each(|(j, x)| {
+            if x == 1 {
+                uf.connect(i, j);
+            }
+        })
+    });
+    uf.count() as i32
+}
+
+/// [684. 冗余连接](https://leetcode.cn/problems/redundant-connection/)
+pub fn find_redundant_connection(edges: Vec<Vec<i32>>) -> Vec<i32> {
+    let mut uf = UnionFind::new(edges.len()+1);
+    let mut curr_count = uf.count();
+    let mut last_edge = vec![];
+
+    for edge in edges {
+        if let &[p, q] = edge.as_slice() {
+            uf.connect(p as usize, q as usize);
+            if uf.count() == curr_count {
+                last_edge = edge.clone();
+            }
+            curr_count = uf.count;
+        }
+    }
+    last_edge
+}
+
+#[cfg(test)]
+mod test {
+    use super::*;
+    use crate::vec2;
+
+    #[test]
+    fn test_find_redundant_connection() {
+        struct Testcase {
+            edges: Vec<Vec<i32>>,
+            expect: Vec<i32>,
+        }
+
+        vec![
+            Testcase {
+                edges: vec2![[1, 2], [1, 3], [2, 3]],
+                expect: vec![2, 3],
+            },
+            Testcase {
+                edges: vec2![[1, 2], [2, 3], [3, 4], [1, 4], [1, 5]],
+                expect: vec![1, 4],
+            },
+        ]
+        .into_iter()
+        .enumerate()
+        .for_each(|(idx, testcase)| {
+            let Testcase { edges, expect } = testcase;
+            let actual = find_redundant_connection(edges);
+            assert_eq!(expect, actual, "case {} failed", idx);
+        });
+    }
+
+    #[test]
+    fn test_find_circle_num() {
+        struct Testcase {
+            is_connected: Vec<Vec<i32>>,
+            expect: i32,
+        }
+
+        vec![
+            Testcase {
+                is_connected: vec2![[1, 1, 0], [1, 1, 0], [0, 0, 1]],
+                expect: 2,
+            },
+            Testcase {
+                is_connected: vec2![[1, 0, 0], [0, 1, 0], [0, 0, 1]],
+                expect: 3,
+            },
+        ]
+        .into_iter()
+        .enumerate()
+        .for_each(|(idx, testcase)| {
+            let Testcase {
+                is_connected,
+                expect,
+            } = testcase;
+            let actual = find_circle_num(is_connected);
+            assert_eq!(expect, actual, "case {} failed", idx);
+        });
+    }
+}

src/array/ext/math.rs

@@ -1,4 +1,12 @@
-//! 数学相关题目
+//! # 数学相关题目
+//!
+//! ## 题目
+//! * 简单
+//!     * [1470. 重新排列数组](shuffle)
+//!     * [1582. 二进制矩阵中的特殊位置](num_special)
+//! * 中等
+//!     * [462. 最小操作次数使数组元素相等 II](min_moves2)
+//!     * [667. 优美的排列 II](construct_array)
 
 /// [462. 最少移动次数使数组元素相等 II](https://leetcode.cn/problems/minimum-moves-to-equal-array-elements-ii/)
 ///
@@ -126,10 +134,118 @@ pub fn construct_array(n: i32, k: i32) -> Vec<i32> {
     result
 }
 
+/// [412. Fizz Buzz](https://leetcode.cn/problems/fizz-buzz/)
+pub fn fizz_buzz(n: i32) -> Vec<String> {
+    (1..=n)
+        .into_iter()
+        .map(|num| {
+            if num % 3 == 0 && num % 5 == 0 {
+                "FizzBuzz".to_string()
+            } else if num % 5 == 0 {
+                "Buzz".to_string()
+            } else if num % 3 == 0 {
+                "Fizz".to_string()
+            } else {
+                num.to_string()
+            }
+        })
+        .collect()
+}
+
+/// [1342. 将数字变成 0 的操作次数](https://leetcode.cn/problems/number-of-steps-to-reduce-a-number-to-zero/)
+///
+/// 思路1: 模拟
+/// ```
+/// pub fn number_of_steps(num: i32) -> i32 {
+///     let mut num = num;
+///     let mut cnt = 0;
+///     while num > 0 {
+///         if num % 2 == 0 {
+///             num = num / 2;
+///         } else {
+///             num = num - 1;
+///         }
+///         cnt += 1;
+///     }
+///     cnt
+/// }
+/// ```
+///
+/// 思路2: 计算
+/// 将num用用二进制表示, 则每次减1， 实际对应为 低位的 1变0， 每次除2, 实际对应为 整体右移一位
+/// 也就是总的操作数为可以视为 将最高位1变为0 的步数
+///
+/// 其中右移总共 `31 - (leading_zero)` (有符号数, 第一位为符号位, 忽略)
+/// 减1 次数, 即为1的个数
+/// 
+/// 注意: 如果原本为0， 这时 `31 - (leading_zero)`可能有溢出
+///
+pub fn number_of_steps(num: i32) -> i32 {
+    (num.count_ones() + 31u32.checked_sub(num.leading_zeros()).unwrap_or(0)) as i32
+}
+
 #[cfg(test)]
 mod tests {
     use super::*;
 
+    #[test]
+    fn test_number_of_steps() {
+        struct TestCase {
+            num: i32,
+            expect: i32,
+        }
+
+        vec![
+            TestCase { num: 14, expect: 6 },
+            TestCase { num: 8, expect: 4 },
+            TestCase {
+                num: 123,
+                expect: 12,
+            },
+            TestCase{num: 0, expect: 0}
+        ]
+        .into_iter()
+        .enumerate()
+        .for_each(|(idx, testcase)| {
+            let TestCase { num, expect } = testcase;
+            let actual = number_of_steps(num);
+            assert_eq!(expect, actual, "case {} failed", idx);
+        });
+    }
+
+    #[test]
+    fn test_fizz_buzz() {
+        struct TestCase {
+            n: i32,
+            expect: Vec<&'static str>,
+        }
+
+        vec![
+            TestCase {
+                n: 3,
+                expect: vec!["1", "2", "Fizz"],
+            },
+            TestCase {
+                n: 5,
+                expect: vec!["1", "2", "Fizz", "4", "Buzz"],
+            },
+            TestCase {
+                n: 15,
+                expect: vec![
+                    "1", "2", "Fizz", "4", "Buzz", "Fizz", "7", "8", "Fizz", "Buzz", "11", "Fizz",
+                    "13", "14", "FizzBuzz",
+                ],
+            },
+        ]
+        .into_iter()
+        .enumerate()
+        .for_each(|(idx, testcase)| {
+            let TestCase { n, expect } = testcase;
+            let actual = fizz_buzz(n);
+            assert_eq!(expect, actual, "case {} failed", idx);
+        });
+    }
+
     #[test]
     fn test_construct_array() {
         struct TestCase {

src/array/ext/quick.rs

@@ -12,10 +12,12 @@
 /// **快排** 主要思路是 分治, 分治可以用递归实现.
 /// 步骤: 1. 确认 *轴*; 2. 根据 *轴*, 将数组切分; 3. 根据轴的位置, 递归处理每个新数组
 /// 步骤一可变性不强, 可以选开头, 结尾, 中间, 随机, 方式很多不过没啥难度和技巧; 步骤三同理
-/// 步骤2, 将小于x的放在x左边, 将大于x的放x右边.
+/// 
+/// 步骤2, 将小于x的放在x左边, 将大于x的放x右边, 有下面两种方式
 ///
-/// 方式1: 使用额外数组, 挑数, 挑出来, 然后拼接在一起
-/// 方式2: 快慢指针
+/// * 方式1: 使用额外数组, 挑数, 挑出来, 然后拼接在一起
+/// * 方式2: 快慢指针
+/// 
 /// ```rust
 /// fn partition(nums: &mut [i32], left: usize, right: usize, pivot: usize) -> usize {
 ///     nums.swap(pivot, right);
@@ -51,7 +53,7 @@
 ///     return r;
 /// }
 /// ```
-/// 不过思路和方式2的快慢指针是一样的, 只是用加了预处理, 将左右边界上明显不符合交换的元素, 做了跳过
+/// 不过思路和方式2的快慢指针是一样的, 只是加了预处理, 将左右边界上明显不符合交换的元素, 做了跳过
 ///
 pub fn sort_array(nums: Vec<i32>) -> Vec<i32> {
     fn partition(nums: &mut [i32], left: usize, right: usize, pivot: usize) -> usize {

src/array/mod.rs

@@ -2,9 +2,11 @@
 //! 线性表相关的都会涉及
 //! 比如数组(连续存储), 简单链表, 字符串(连续存储)
 
+/// vec2![[1,2,3],[3,2,1]] => vec![vec![1,2,3], vec![3,2,1]]
+
 /// 一些周边题目
 pub mod ext;
+/// 暂时不归类的题目
+pub mod no_class;
 /// 姑且能所做一个系列的题目
 pub mod ser;
-/// 暂时不归类的题目
-pub mod no_class;