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solve leetcode |993|[Cousins In Binary Tree]
1 parent 818297f commit 27dc843

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+235
-174
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‎.cargo/config.toml.example

-14
This file was deleted.

‎README.md

+1
Original file line numberDiff line numberDiff line change
@@ -356,6 +356,7 @@ leetcode_solutions
356356
|976|[Largest Perimeter Triangle](https://leetcode.com/problems/largest-perimeter-triangle/)|[Rust](src/easy/leetcode_easy.rs)|
357357
|977|[Squares of a Sorted Array](https://leetcode.com/problems/squares-of-a-sorted-array/)|[Rust](src/easy/array/squares_of_a_sorted_array.rs)|
358358
|989|[Add to Array-Form of Integer](https://leetcode.com/problems/add-to-array-form-of-integer/)|[Rust](src/easy/leetcode_very_easy.rs)|
359+
|993|[Cousins In Binary Tree](https://leetcode.com/problems/cousins-in-binary-tree/)|[Rust](src/binary_tree/level_order_traversal.rs)|
359360
|1002|[Find Common Characters](https://leetcode.com/problems/find-common-characters/)|[Rust](src/easy/string/find_common_characters.rs)|
360361
|1006|[Clumsy Factorial](https://leetcode.com/problems/clumsy-factorial/)|[Rust](src/eval_math_expression/mod.rs)|
361362
|1018|[Binary Prefix Divisible By 5](https://leetcode.com/problems/binary-prefix-divisible-by-5/)|[Rust](src/easy/leetcode_very_easy.rs)|

‎src/binary_search/median_of_two_sorted_arrays.rs

+76-89
Original file line numberDiff line numberDiff line change
@@ -1,107 +1,93 @@
11
//! https://leetcode.com/problems/median-of-two-sorted-arrays/
22
//! TODO 这题可读性最好最容易记的应该是find_kth递归的解法，我写的Rust解法太长了可读性并不好，建议看我写的Python解法版本
33
4-
struct Solution;
5-
6-
impl Solution {
7-
/// [4ms, O(n*logn)]即便用了两个数组合并后排序的完全没利用上两个数组已经有序的笨方法，Rust的性能还是能跑进4ms
8-
fn my_brute_force(mut nums1: Vec<i32>, mut nums2: Vec<i32>) -> f64 {
9-
if nums1.len() < nums2.len() {
10-
return Self::my_brute_force(nums2, nums1);
11-
}
12-
nums1.append(&mut nums2);
13-
nums1.sort_unstable();
14-
let len = nums1.len();
15-
if len % 2 == 0 {
16-
(nums1[len / 2 - 1] + nums1[len / 2]) as f64 / 2_f64
17-
} else {
18-
f64::from(nums1[len / 2])
19-
}
4+
/// [4ms, O(n*logn)]即便用了两个数组合并后排序的完全没利用上两个数组已经有序的笨方法，Rust的性能还是能跑进4ms
5+
fn my_brute_force(mut nums1: Vec<i32>, mut nums2: Vec<i32>) -> f64 {
6+
if nums1.len() < nums2.len() {
7+
return my_brute_force(nums2, nums1);
208
}
9+
nums1.append(&mut nums2);
10+
nums1.sort_unstable();
11+
let len = nums1.len();
12+
if len % 2 == 0 {
13+
(nums1[len / 2 - 1] + nums1[len / 2]) as f64 / 2_f64
14+
} else {
15+
f64::from(nums1[len / 2])
16+
}
17+
}
2118

22-
/// [0ms, O(n)]既然两个数组已经有序，那么可以用归并排序的归并操作去合并数组提升性能，使用二分法能达到更快的logn时间复杂度
23-
fn merge_sort_solution(nums1: Vec<i32>, nums2: Vec<i32>) -> f64 {
24-
let (m, n, mut i, mut j) = (nums1.len(), nums2.len(), 0_usize, 0_usize);
25-
let len = m + n;
26-
let half_len = len / 2;
27-
let mut merged = Vec::with_capacity(len);
28-
while i < m && j < n {
29-
if nums1[i] <= nums2[j] {
30-
merged.push(nums1[i]);
31-
i += 1;
32-
} else {
33-
merged.push(nums2[j]);
34-
j += 1;
35-
}
36-
}
37-
38-
// 如果元素个数已经够了，就提前返回，提升性能
39-
if merged.len() > half_len {
40-
return if len % 2 == 0 {
41-
(merged[half_len - 1] + merged[half_len]) as f64 / 2_f64
42-
} else {
43-
merged[half_len] as f64
44-
};
45-
}
46-
47-
while i < m {
19+
/// [0ms, O(n)]既然两个数组已经有序，那么可以用归并排序的归并操作去合并数组提升性能，使用二分法能达到更快的logn时间复杂度
20+
fn merge_sort_solution(nums1: Vec<i32>, nums2: Vec<i32>) -> f64 {
21+
let (m, n, mut i, mut j) = (nums1.len(), nums2.len(), 0_usize, 0_usize);
22+
let len = m + n;
23+
let half_len = len / 2;
24+
let mut merged = Vec::with_capacity(len);
25+
while i < m && j < n {
26+
if nums1[i] <= nums2[j] {
4827
merged.push(nums1[i]);
4928
i += 1;
50-
}
51-
while j < n {
29+
} else {
5230
merged.push(nums2[j]);
5331
j += 1;
5432
}
55-
if len % 2 == 0 {
33+
}
34+
35+
// 如果元素个数已经够了，就提前返回，提升性能
36+
if merged.len() > half_len {
37+
return if len % 2 == 0 {
5638
(merged[half_len - 1] + merged[half_len]) as f64 / 2_f64
5739
} else {
5840
merged[half_len] as f64
59-
}
41+
};
6042
}
61-
}
6243

63-
#[cfg(test)]
64-
mod test_find_median_sorted_arrays {
65-
use super::Solution;
66-
67-
const TEST_CASES: [(&[i32], &[i32], f64); 17] = [
68-
(&[1, 2, 3, 4], &[3, 6, 8, 9], 3.5),
69-
(&[1, 5, 6, 7], &[2, 3, 4, 8], 4.5),
70-
(&[1, 2], &[3, 4, 5, 6, 7], 4f64),
71-
(&[4, 5, 6], &[1, 2, 3], 3.5),
72-
(&[4, 5], &[1, 2, 3, 6], 3.5),
73-
(&[1, 3], &[2, 4, 5, 6], 3.5),
74-
(&[1, 2], &[3, 4, 5, 6], 3.5),
75-
(&[1, 3], &[2, 4, 5], 3f64),
76-
(&[1, 2, 3], &[4, 5], 3f64),
77-
(&[3, 4], &[1, 2, 5], 3f64),
78-
(&[-2, -1], &[3], -1f64),
79-
(&[1, 3], &[2], 2_f64),
80-
(&[3], &[-2, -1], -1f64),
81-
(&[1, 2], &[3, 4], 2.5),
82-
(&[4, 5], &[1, 2, 3], 3f64),
83-
(&[1, 2], &[1, 2, 3], 2_f64),
84-
(&[1, 2, 3], &[1, 2, 3], 2_f64),
85-
];
44+
while i < m {
45+
merged.push(nums1[i]);
46+
i += 1;
47+
}
48+
while j < n {
49+
merged.push(nums2[j]);
50+
j += 1;
51+
}
52+
if len % 2 == 0 {
53+
(merged[half_len - 1] + merged[half_len]) as f64 / 2_f64
54+
} else {
55+
merged[half_len] as f64
56+
}
57+
}
58+
const TEST_CASES: [(&[i32], &[i32], f64); 17] = [
59+
(&[1, 2, 3, 4], &[3, 6, 8, 9], 3.5),
60+
(&[1, 5, 6, 7], &[2, 3, 4, 8], 4.5),
61+
(&[1, 2], &[3, 4, 5, 6, 7], 4_f64),
62+
(&[4, 5, 6], &[1, 2, 3], 3.5),
63+
(&[4, 5], &[1, 2, 3, 6], 3.5),
64+
(&[1, 3], &[2, 4, 5, 6], 3.5),
65+
(&[1, 2], &[3, 4, 5, 6], 3.5),
66+
(&[1, 3], &[2, 4, 5], 3_f64),
67+
(&[1, 2, 3], &[4, 5], 3_f64),
68+
(&[3, 4], &[1, 2, 5], 3_f64),
69+
(&[-2, -1], &[3], -1_f64),
70+
(&[1, 3], &[2], 2_f64),
71+
(&[3], &[-2, -1], -1_f64),
72+
(&[1, 2], &[3, 4], 2.5),
73+
(&[4, 5], &[1, 2, 3], 3_f64),
74+
(&[1, 2], &[1, 2, 3], 2_f64),
75+
(&[1, 2, 3], &[1, 2, 3], 2_f64),
76+
];
8677

87-
#[test]
88-
fn test_my_brute_force() {
89-
for &(nums1, nums2, expected) in TEST_CASES.iter() {
90-
assert_eq!(
91-
Solution::my_brute_force(nums1.to_vec(), nums2.to_vec()),
92-
expected
93-
);
94-
}
78+
#[test]
79+
fn test_my_brute_force() {
80+
for (nums1, nums2, expected) in TEST_CASES {
81+
assert!((my_brute_force(nums1.to_vec(), nums2.to_vec()) - expected).abs() < f64::EPSILON);
9582
}
83+
}
9684

97-
#[test]
98-
fn test_merge_sort_solution() {
99-
for &(nums1, nums2, expected) in TEST_CASES.iter() {
100-
assert_eq!(
101-
Solution::merge_sort_solution(nums1.to_vec(), nums2.to_vec()),
102-
expected
103-
);
104-
}
85+
#[test]
86+
fn test_merge_sort_solution() {
87+
for (nums1, nums2, expected) in TEST_CASES {
88+
assert!(
89+
(merge_sort_solution(nums1.to_vec(), nums2.to_vec()) - expected).abs() < f64::EPSILON
90+
);
10591
}
10692
}
10793

@@ -198,9 +184,10 @@ fn find_median_sorted_arrays(nums1: Vec<i32>, nums2: Vec<i32>) -> f64 {
198184
f64::from(nums1[len_a - 1] + nums2[0]) / 2_f64
199185
} else {
200186
// [1,3]、[2,4,5,6]
201-
f64::from(nums2[b_divider_right_index - 2].max(nums1[len_a - 1])
202-
+ nums2[b_divider_right_index - 1])
203-
/ 2_f64
187+
f64::from(
188+
nums2[b_divider_right_index - 2].max(nums1[len_a - 1])
189+
+ nums2[b_divider_right_index - 1],
190+
) / 2_f64
204191
}
205192
} else {
206193
// [1,2]、[3,4,5,6,7]

‎src/binary_tree/level_order_traversal.rs

+65
Original file line numberDiff line numberDiff line change
@@ -94,3 +94,68 @@ fn zigzag_level_order(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<Vec<i32>> {
9494
}
9595
ret
9696
}
97+
98+
/// https://leetcode.com/problems/cousins-in-binary-tree/
99+
/// 如果二叉树的两个节点深度相同(处于同一层)，但**父节点不同**，则它们是一对堂兄弟节点
100+
/// 需要知道每个节点的三个信息: 层数、值、父节点的值
101+
fn is_cousins(root: Option<Rc<RefCell<TreeNode>>>, x: i32, y: i32) -> bool {
102+
// (depth, parent)
103+
let mut node_x: Option<(u8, i32)> = None;
104+
let mut node_y: Option<(u8, i32)> = None;
105+
let mut queue = std::collections::VecDeque::new();
106+
queue.push_back((root, 0)); // (cur_node, cur_parent)
107+
// add sentinel node to queue end
108+
queue.push_back((None, 0));
109+
let mut cur_depth = 0_u8;
110+
while let Some((cur_node, cur_parent)) = queue.pop_front() {
111+
if let Some(cur_node) = cur_node {
112+
let cur_node = cur_node.borrow();
113+
114+
if cur_node.val == x {
115+
if let Some((node_y_depth, node_y_parent)) = node_y {
116+
return cur_depth == node_y_depth && cur_parent != node_y_parent;
117+
} else {
118+
node_x = Some((cur_depth, cur_parent));
119+
}
120+
} else if cur_node.val == y {
121+
if let Some((node_x_depth, node_x_parent)) = node_x {
122+
return cur_depth == node_x_depth && cur_parent != node_x_parent;
123+
} else {
124+
node_y = Some((cur_depth, cur_parent));
125+
}
126+
}
127+
128+
if cur_node.left.is_some() {
129+
queue.push_back((cur_node.left.clone(), cur_node.val));
130+
}
131+
if cur_node.right.is_some() {
132+
queue.push_back((cur_node.right.clone(), cur_node.val));
133+
}
134+
} else {
135+
cur_depth += 1;
136+
// add level separator to queue end
137+
if !queue.is_empty() {
138+
queue.push_back((None, 0));
139+
}
140+
}
141+
}
142+
false
143+
}
144+
145+
#[test]
146+
fn test_is_cousins() {
147+
#[allow(non_upper_case_globals)]
148+
const null: i32 = TreeNode::NULL;
149+
const TEST_CASES: [(&[i32], i32, i32, bool); 3] = [
150+
(&[1, 2, 3, 4], 4, 3, false),
151+
(&[1, 2, 3, null, 4, null, 5], 5, 4, true),
152+
(&[1, 2, 3, null, 4], 2, 3, false),
153+
];
154+
for (root, x, y, expected) in TEST_CASES {
155+
let root = super::deserialize_vec_to_binary_tree(root);
156+
println!("{}", "=".repeat(20));
157+
super::print_binary_tree(root.clone()).unwrap();
158+
println!("x={}, y={}, expected={}", x, y, expected);
159+
assert_eq!(is_cousins(root, x, y), expected);
160+
}
161+
}

‎src/counter/mod.rs

+2-2
Original file line numberDiff line numberDiff line change
@@ -58,8 +58,8 @@ fn unique_occurrences(arr: Vec<i32>) -> bool {
5858

5959
#[test]
6060
fn test_unique_occurrences() {
61-
assert_eq!(unique_occurrences(vec![1, 2, 2, 1, 1, 3]), true);
62-
assert_eq!(unique_occurrences(vec![1, 2]), false);
61+
assert!(unique_occurrences(vec![1, 2, 2, 1, 1, 3]));
62+
assert!(!unique_occurrences(vec![1, 2]));
6363
}
6464

6565
/// https://leetcode.com/problems/how-many-numbers-are-smaller-than-the-current-number/

‎src/dp/drop_eggs.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -105,7 +105,7 @@ const TEST_CASES: [(i32, i32, i32); 2] = [
105105

106106
#[test]
107107
fn test() {
108-
for &(eggs_k, n, times) in TEST_CASES.iter() {
108+
for (eggs_k, n, times) in TEST_CASES {
109109
assert_eq!(dp_binary_search(eggs_k, n), times);
110110
}
111111
}

‎src/dp/freedom_trail.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -94,7 +94,7 @@ const TEST_CASES: [(&str, &str, i32); 1] = [("godding", "gd", 4)];
9494

9595
#[test]
9696
fn test_find_rotate_steps() {
97-
for &(ring, key, steps) in TEST_CASES.iter() {
97+
for (ring, key, steps) in TEST_CASES {
9898
assert_eq!(find_rotate_steps(ring.to_owned(), key.to_owned()), steps);
9999
assert_eq!(
100100
find_rotate_steps_optimized(ring.to_owned(), key.to_owned()),

‎src/dp/triangle.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -39,7 +39,7 @@ fn from_bottom_to_top(mut triangle: Vec<Vec<i32>>) -> i32 {
3939
#[test]
4040
fn test() {
4141
let test_cases = vec![(vec_vec![[2], [3, 4], [6, 5, 7], [4, 1, 8, 3]], 11)];
42-
for (input, output) in test_cases.into_iter() {
42+
for (input, output) in test_cases {
4343
assert_eq!(from_top_to_bottom(input.clone()), output);
4444
assert_eq!(from_bottom_to_top(input), output);
4545
}

‎src/dp/unique_paths.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -69,7 +69,7 @@ fn unique_paths(m: i32, n: i32) -> i32 {
6969
#[test]
7070
fn test_unique_paths() {
7171
const TEST_CASES: [(i32, i32, i32); 2] = [(23, 12, 193536720), (51, 9, 1916797311)];
72-
for &(m, n, expected) in TEST_CASES.iter() {
72+
for (m, n, expected) in TEST_CASES {
7373
assert_eq!(unique_paths(m, n), expected);
7474
}
7575
}

‎src/easy/array/squares_of_a_sorted_array.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -35,7 +35,7 @@ impl Solution {
3535
fn test() {
3636
const TEST_CASES: [&[i32]; 2] = [&[-4, -1, 0, 3, 10], &[-7, -3, 2, 3, 11]];
3737

38-
for &nums in TEST_CASES.iter() {
38+
for nums in TEST_CASES {
3939
assert!(Solution::sorted_squares(nums.to_vec()).is_sorted());
4040
assert!(Solution::two_pointers_sorted_squares(nums.to_vec()).is_sorted());
4141
}

‎src/easy/grid_or_matrix/island_perimeter.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -39,7 +39,7 @@ fn test_island_perimeter() {
3939
vec_vec![[0, 1, 0, 0], [1, 1, 1, 0], [0, 1, 0, 0], [1, 1, 0, 0]],
4040
16,
4141
)];
42-
for (grid, perimeter) in test_cases.into_iter() {
42+
for (grid, perimeter) in test_cases {
4343
assert_eq!(island_perimeter(grid), perimeter);
4444
}
4545
}

‎src/easy/grid_or_matrix/matrix_diagonal_traverse.rs

+5-5
Original file line numberDiff line numberDiff line change
@@ -30,8 +30,8 @@ fn diagonal_traverse_from_bottom_left_in_bottom_right_direction() {
3030
}
3131
}
3232
// bottom_right_direction traverse
33-
for i in 0..m {
34-
println!("{:?}", mat[i]);
33+
for row in mat.into_iter().take(m) {
34+
println!("{:?}", row);
3535
}
3636
}
3737

@@ -111,8 +111,8 @@ fn diagonal_traverse_from_top_left_in_bottom_left_direction() {
111111
order += 1;
112112
}
113113
}
114-
for i in 0..m {
115-
println!("{:?}", mat[i]);
114+
for row in mat {
115+
println!("{:?}", row);
116116
}
117117
}
118118

@@ -166,7 +166,7 @@ fn test_find_diagonal_order() {
166166
),
167167
(vec_vec![[2, 3], [2, 3]], vec![2, 3]),
168168
];
169-
for (input, output) in test_cases.into_iter() {
169+
for (input, output) in test_cases {
170170
assert_eq!(find_diagonal_order(input), output);
171171
}
172172
}

‎src/easy/grid_or_matrix/rotate_matrix.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -42,7 +42,7 @@ fn test_rotate() {
4242
vec_vec![[1, 2, 3], [4, 5, 6], [7, 8, 9]],
4343
vec_vec![[7, 4, 1], [8, 5, 2], [9, 6, 3]],
4444
)];
45-
for (mut input, output) in test_cases.into_iter() {
45+
for (mut input, output) in test_cases {
4646
rotate(&mut input);
4747
assert_eq!(input, output);
4848
}

‎src/easy/grid_or_matrix/spiral_matrix.rs

+1-1
Original file line numberDiff line numberDiff line change
@@ -63,7 +63,7 @@ fn test_spiral_matrix_1() {
6363
vec![1, 2, 3, 4, 8, 12, 11, 10, 9, 5, 6, 7],
6464
),
6565
];
66-
for (input, output) in test_cases.into_iter() {
66+
for (input, output) in test_cases {
6767
assert_eq!(spiral_matrix_1(input), output);
6868
}
6969
}

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