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+4957
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lines changed

‎.editorconfig

+7
Original file line numberDiff line numberDiff line change
@@ -0,0 +1,7 @@
1+
root = true
2+
3+
[*]
4+
charset = utf-8
5+
end_of_line = lf
6+
insert_final_newline = true
7+
trim_trailing_whitespace = true

‎solutions/3100-3199/3157.find-the-level-of-tree-with-minimum-sum/Solution.cpp

+37-37
Original file line numberDiff line numberDiff line change
@@ -1,38 +1,38 @@
1-
/**
2-
* Definition for a binary tree node.
3-
* struct TreeNode {
4-
* int val;
5-
* TreeNode *left;
6-
* TreeNode *right;
7-
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
8-
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
9-
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10-
* };
11-
*/
12-
class Solution {
13-
public:
14-
int minimumLevel(TreeNode* root) {
15-
queue<TreeNode*> q{{root}};
16-
int ans = 0;
17-
long long s = 1LL << 60;
18-
for (int level = 1; q.size(); ++level) {
19-
long long t = 0;
20-
for (int m = q.size(); m; --m) {
21-
TreeNode* node = q.front();
22-
q.pop();
23-
t += node->val;
24-
if (node->left) {
25-
q.push(node->left);
26-
}
27-
if (node->right) {
28-
q.push(node->right);
29-
}
30-
}
31-
if (s > t) {
32-
s = t;
33-
ans = level;
34-
}
35-
}
36-
return ans;
37-
}
1+
/**
2+
* Definition for a binary tree node.
3+
* struct TreeNode {
4+
* int val;
5+
* TreeNode *left;
6+
* TreeNode *right;
7+
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
8+
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
9+
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
10+
* };
11+
*/
12+
class Solution {
13+
public:
14+
int minimumLevel(TreeNode* root) {
15+
queue<TreeNode*> q{{root}};
16+
int ans = 0;
17+
long long s = 1LL << 60;
18+
for (int level = 1; q.size(); ++level) {
19+
long long t = 0;
20+
for (int m = q.size(); m; --m) {
21+
TreeNode* node = q.front();
22+
q.pop();
23+
t += node->val;
24+
if (node->left) {
25+
q.push(node->left);
26+
}
27+
if (node->right) {
28+
q.push(node->right);
29+
}
30+
}
31+
if (s > t) {
32+
s = t;
33+
ans = level;
34+
}
35+
}
36+
return ans;
37+
}
3838
};

‎solutions/3100-3199/3157.find-the-level-of-tree-with-minimum-sum/Solution.go

+30-30
Original file line numberDiff line numberDiff line change
@@ -1,31 +1,31 @@
1-
/**
2-
* Definition for a binary tree node.
3-
* type TreeNode struct {
4-
* Val int
5-
* Left *TreeNode
6-
* Right *TreeNode
7-
* }
8-
*/
9-
func minimumLevel(root *TreeNode) (ans int) {
10-
q := []*TreeNode{root}
11-
s := math.MaxInt64
12-
for level := 1; len(q) > 0; level++ {
13-
t := 0
14-
for m := len(q); m > 0; m-- {
15-
node := q[0]
16-
q = q[1:]
17-
t += node.Val
18-
if node.Left != nil {
19-
q = append(q, node.Left)
20-
}
21-
if node.Right != nil {
22-
q = append(q, node.Right)
23-
}
24-
}
25-
if s > t {
26-
s = t
27-
ans = level
28-
}
29-
}
30-
return
1+
/**
2+
* Definition for a binary tree node.
3+
* type TreeNode struct {
4+
* Val int
5+
* Left *TreeNode
6+
* Right *TreeNode
7+
* }
8+
*/
9+
func minimumLevel(root *TreeNode) (ans int) {
10+
q := []*TreeNode{root}
11+
s := math.MaxInt64
12+
for level := 1; len(q) > 0; level++ {
13+
t := 0
14+
for m := len(q); m > 0; m-- {
15+
node := q[0]
16+
q = q[1:]
17+
t += node.Val
18+
if node.Left != nil {
19+
q = append(q, node.Left)
20+
}
21+
if node.Right != nil {
22+
q = append(q, node.Right)
23+
}
24+
}
25+
if s > t {
26+
s = t
27+
ans = level
28+
}
29+
}
30+
return
3131
}

‎solutions/3100-3199/3157.find-the-level-of-tree-with-minimum-sum/Solution.java

+40-40
Original file line numberDiff line numberDiff line change
@@ -1,41 +1,41 @@
1-
/**
2-
* Definition for a binary tree node.
3-
* public class TreeNode {
4-
* int val;
5-
* TreeNode left;
6-
* TreeNode right;
7-
* TreeNode() {}
8-
* TreeNode(int val) { this.val = val; }
9-
* TreeNode(int val, TreeNode left, TreeNode right) {
10-
* this.val = val;
11-
* this.left = left;
12-
* this.right = right;
13-
* }
14-
* }
15-
*/
16-
class Solution {
17-
public int minimumLevel(TreeNode root) {
18-
Deque<TreeNode> q = new ArrayDeque<>();
19-
q.offer(root);
20-
int ans = 0;
21-
long s = Long.MAX_VALUE;
22-
for (int level = 1; !q.isEmpty(); ++level) {
23-
long t = 0;
24-
for (int m = q.size(); m > 0; --m) {
25-
TreeNode node = q.poll();
26-
t += node.val;
27-
if (node.left != null) {
28-
q.offer(node.left);
29-
}
30-
if (node.right != null) {
31-
q.offer(node.right);
32-
}
33-
}
34-
if (s > t) {
35-
s = t;
36-
ans = level;
37-
}
38-
}
39-
return ans;
40-
}
1+
/**
2+
* Definition for a binary tree node.
3+
* public class TreeNode {
4+
* int val;
5+
* TreeNode left;
6+
* TreeNode right;
7+
* TreeNode() {}
8+
* TreeNode(int val) { this.val = val; }
9+
* TreeNode(int val, TreeNode left, TreeNode right) {
10+
* this.val = val;
11+
* this.left = left;
12+
* this.right = right;
13+
* }
14+
* }
15+
*/
16+
class Solution {
17+
public int minimumLevel(TreeNode root) {
18+
Deque<TreeNode> q = new ArrayDeque<>();
19+
q.offer(root);
20+
int ans = 0;
21+
long s = Long.MAX_VALUE;
22+
for (int level = 1; !q.isEmpty(); ++level) {
23+
long t = 0;
24+
for (int m = q.size(); m > 0; --m) {
25+
TreeNode node = q.poll();
26+
t += node.val;
27+
if (node.left != null) {
28+
q.offer(node.left);
29+
}
30+
if (node.right != null) {
31+
q.offer(node.right);
32+
}
33+
}
34+
if (s > t) {
35+
s = t;
36+
ans = level;
37+
}
38+
}
39+
return ans;
40+
}
4141
}

‎solutions/3100-3199/3157.find-the-level-of-tree-with-minimum-sum/Solution.py

+25-25
Original file line numberDiff line numberDiff line change
@@ -1,25 +1,25 @@
1-
# Definition for a binary tree node.
2-
# class TreeNode:
3-
# def __init__(self, val=0, left=None, right=None):
4-
# self.val = val
5-
# self.left = left
6-
# self.right = right
7-
class Solution:
8-
def minimumLevel(self, root: Optional[TreeNode]) -> int:
9-
q = deque([root])
10-
ans = 0
11-
level, s = 1, inf
12-
while q:
13-
t = 0
14-
for _ in range(len(q)):
15-
node = q.popleft()
16-
t += node.val
17-
if node.left:
18-
q.append(node.left)
19-
if node.right:
20-
q.append(node.right)
21-
if s > t:
22-
s = t
23-
ans = level
24-
level += 1
25-
return ans
1+
# Definition for a binary tree node.
2+
# class TreeNode:
3+
# def __init__(self, val=0, left=None, right=None):
4+
# self.val = val
5+
# self.left = left
6+
# self.right = right
7+
class Solution:
8+
def minimumLevel(self, root: Optional[TreeNode]) -> int:
9+
q = deque([root])
10+
ans = 0
11+
level, s = 1, inf
12+
while q:
13+
t = 0
14+
for _ in range(len(q)):
15+
node = q.popleft()
16+
t += node.val
17+
if node.left:
18+
q.append(node.left)
19+
if node.right:
20+
q.append(node.right)
21+
if s > t:
22+
s = t
23+
ans = level
24+
level += 1
25+
return ans

‎solutions/3100-3199/3158.find-the-xor-of-numbers-which-appear-twice/Solution2.cpp

+15-15
Original file line numberDiff line numberDiff line change
@@ -1,15 +1,15 @@
1-
class Solution {
2-
public:
3-
int duplicateNumbersXOR(vector<int>& nums) {
4-
int ans = 0;
5-
long long mask = 0;
6-
for (int x : nums) {
7-
if (mask >> x & 1) {
8-
ans ^= x;
9-
} else {
10-
mask |= 1LL << x;
11-
}
12-
}
13-
return ans;
14-
}
15-
};
1+
class Solution {
2+
public:
3+
int duplicateNumbersXOR(vector<int>& nums) {
4+
int ans = 0;
5+
long long mask = 0;
6+
for (int x : nums) {
7+
if (mask >> x & 1) {
8+
ans ^= x;
9+
} else {
10+
mask |= 1LL << x;
11+
}
12+
}
13+
return ans;
14+
}
15+
};

‎solutions/3100-3199/3158.find-the-xor-of-numbers-which-appear-twice/Solution2.go

+11-11
Original file line numberDiff line numberDiff line change
@@ -1,11 +1,11 @@
1-
func duplicateNumbersXOR(nums []int) (ans int) {
2-
mask := 0
3-
for _, x := range nums {
4-
if mask>>x&1 == 1 {
5-
ans ^= x
6-
} else {
7-
mask |= 1 << x
8-
}
9-
}
10-
return
11-
}
1+
func duplicateNumbersXOR(nums []int) (ans int) {
2+
mask := 0
3+
for _, x := range nums {
4+
if mask>>x&1 == 1 {
5+
ans ^= x
6+
} else {
7+
mask |= 1 << x
8+
}
9+
}
10+
return
11+
}

‎solutions/3100-3199/3158.find-the-xor-of-numbers-which-appear-twice/Solution2.java

+14-14
Original file line numberDiff line numberDiff line change
@@ -1,14 +1,14 @@
1-
class Solution {
2-
public int duplicateNumbersXOR(int[] nums) {
3-
int ans = 0;
4-
long mask = 0;
5-
for (int x : nums) {
6-
if ((mask >> x & 1) == 1) {
7-
ans ^= x;
8-
} else {
9-
mask |= 1L << x;
10-
}
11-
}
12-
return ans;
13-
}
14-
}
1+
class Solution {
2+
public int duplicateNumbersXOR(int[] nums) {
3+
int ans = 0;
4+
long mask = 0;
5+
for (int x : nums) {
6+
if ((mask >> x & 1) == 1) {
7+
ans ^= x;
8+
} else {
9+
mask |= 1L << x;
10+
}
11+
}
12+
return ans;
13+
}
14+
}

‎solutions/3100-3199/3158.find-the-xor-of-numbers-which-appear-twice/Solution2.py

+9-9
Original file line numberDiff line numberDiff line change
@@ -1,9 +1,9 @@
1-
class Solution:
2-
def duplicateNumbersXOR(self, nums: List[int]) -> int:
3-
ans = mask = 0
4-
for x in nums:
5-
if mask >> x & 1:
6-
ans ^= x
7-
else:
8-
mask |= 1 << x
9-
return ans
1+
class Solution:
2+
def duplicateNumbersXOR(self, nums: List[int]) -> int:
3+
ans = mask = 0
4+
for x in nums:
5+
if mask >> x & 1:
6+
ans ^= x
7+
else:
8+
mask |= 1 << x
9+
return ans

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