Huahua's Tech Road

花花酱 LeetCode 35. Search Insert Position

By zxi on January 31, 2020

Given a sorted array and a target value, return the index if the target is found. If not, return the index where it would be if it were inserted in order.

You may assume no duplicates in the array.

Example 1:

Input: [1,3,5,6], 5
Output: 2

Example 2:

Input: [1,3,5,6], 2
Output: 1

Example 3:

Input: [1,3,5,6], 7
Output: 4

Example 4:

Input: [1,3,5,6], 0
Output: 0

Solution: Binary Search

Find the number or upper bound if doesn’t exist.

Time complexity: O(logn)
Space complexity: O1)

C++

// Author: Huahua

class Solution {

public:

int searchInsert(vector<int>& nums, int target) {

int l = 0, r = nums.size();

while (l < r) {

int m = l + (r - l) / 2;

if (nums[m] == target)

return m;

else if (nums[m] > target)

r = m;

else

l = m + 1;

}

return l;

}

};

花花酱 LeetCode 721. Accounts Merge

By zxi on January 29, 2020

Given a list accounts, each element accounts[i] is a list of strings, where the first element accounts[i][0] is a name, and the rest of the elements are emails representing emails of the account.

Now, we would like to merge these accounts. Two accounts definitely belong to the same person if there is some email that is common to both accounts. Note that even if two accounts have the same name, they may belong to different people as people could have the same name. A person can have any number of accounts initially, but all of their accounts definitely have the same name.

After merging the accounts, return the accounts in the following format: the first element of each account is the name, and the rest of the elements are emails in sorted order. The accounts themselves can be returned in any order.

Example 1:

Input: 
accounts = [["John", "johnsmith@mail.com", "john00@mail.com"], ["John", "johnnybravo@mail.com"], ["John", "johnsmith@mail.com", "john_newyork@mail.com"], ["Mary", "mary@mail.com"]]
Output: [["John", 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com'],  ["John", "johnnybravo@mail.com"], ["Mary", "mary@mail.com"]]
Explanation: 
The first and third John's are the same person as they have the common email "johnsmith@mail.com".
The second John and Mary are different people as none of their email addresses are used by other accounts.
We could return these lists in any order, for example the answer [['Mary', 'mary@mail.com'], ['John', 'johnnybravo@mail.com'], 
['John', 'john00@mail.com', 'john_newyork@mail.com', 'johnsmith@mail.com']] would still be accepted.

Note:The length of accounts will be in the range [1, 1000].The length of accounts[i] will be in the range [1, 10].The length of accounts[i][j] will be in the range [1, 30].

Solution: Union-Find

C++

// Author: Huahua

class Solution {

public:

vector<vector<string>> accountsMerge(vector<vector<string>>& accounts) {

unordered_map<string_view, int> ids; // email to id

unordered_map<int, string_view> names; // id to name

vector<int> p(10000);

iota(begin(p), end(p), 0);

function<int(int)> find = [&](int x) {

if (p[x] != x) p[x] = find(p[x]);

return p[x];

};

auto getIdByEmail = [&](string_view email) {

auto it = ids.find(email);

if (it == ids.end()) {

int id = ids.size();

return ids[email] = id;

}

return it->second;

};

for (const auto& account : accounts) {

int u = find(getIdByEmail(account[1]));

for (int i = 2; i < account.size(); ++i)

p[find(u)] = find(getIdByEmail(account[i]));

names[find(u)] = string_view(account[0]);

}

unordered_map<int, set<string>> mergered;

for (const auto& account : accounts)

for (int i = 1; i < account.size(); ++i) {

int id = find(getIdByEmail(account[i]));

mergered[id].insert(account[i]);

}

vector<vector<string>> ans;

for (const auto& kv : mergered) {

ans.push_back({string(names[kv.first])});

ans.back().insert(ans.back().end(), kv.second.begin(), kv.second.end());

}

return ans;

}

};

花花酱 LeetCode 138. Copy List with Random Pointer

By zxi on January 29, 2020

A linked list is given such that each node contains an additional random pointer which could point to any node in the list or null.

Return a deep copy of the list.

The Linked List is represented in the input/output as a list of n nodes. Each node is represented as a pair of [val, random_index] where:

val: an integer representing Node.val
random_index: the index of the node (range from 0 to n-1) where random pointer points to, or null if it does not point to any node.

Example 1:

Input: head = [[7,null],[13,0],[11,4],[10,2],[1,0]]
Output: [[7,null],[13,0],[11,4],[10,2],[1,0]]

Example 2:

Input: head = [[1,1],[2,1]]
Output: [[1,1],[2,1]]

Example 3:

Input: head = [[3,null],[3,0],[3,null]]
Output: [[3,null],[3,0],[3,null]]

Example 4:

Input: head = []
Output: []
Explanation: Given linked list is empty (null pointer), so return null.

Constraints:

-10000 <= Node.val <= 10000
Node.random is null or pointing to a node in the linked list.
Number of Nodes will not exceed 1000.

Solution: Hashtable

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

Node* copyRandomList(Node* head) {

if (!head) return head;

unordered_map<Node*, Node*> m;

Node* cur = m[head] = new Node(head->val);

Node* ans = cur;

while (head) {

if (head->random) {

auto it = m.find(head->random);

if (it == end(m))

it = m.emplace(head->random, new Node(head->random->val)).first;

cur->random = it->second;

}

if (head->next) {

auto it = m.find(head->next);

if (it == end(m))

it = m.emplace(head->next, new Node(head->next->val)).first;

cur->next = it->second;

}

head = head->next;

cur = cur->next;

}

return ans;

}

};

花花酱 LeetCode 235. Lowest Common Ancestor of a Binary Search Tree

By zxi on January 29, 2020

Given a binary search tree (BST), find the lowest common ancestor (LCA) of two given nodes in the BST.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given binary search tree: root = [6,2,8,0,4,7,9,null,null,3,5]

Example 1:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 8
Output: 6
Explanation: The LCA of nodes 2 and 8 is 6.

Example 2:

Input: root = [6,2,8,0,4,7,9,null,null,3,5], p = 2, q = 4
Output: 2
Explanation: The LCA of nodes 2 and 4 is 2, since a node can be a descendant of itself according to the LCA definition.

Note:

All of the nodes’ values will be unique.
p and q are different and both values will exist in the BST.

Solution: Recursion

Time complexity: O(n)
Space complexity: O(n)

C++

// Author: Huahua

class Solution {

public:

TreeNode* lowestCommonAncestor(TreeNode* root, TreeNode* p, TreeNode* q) {

if (p->val < root->val && q->val < root->val)

return lowestCommonAncestor(root->left, p, q);

if (p->val > root->val && q->val > root->val)

return lowestCommonAncestor(root->right, p, q);

return root;

}

};

花花酱 LeetCode 257. Binary Tree Paths

By zxi on January 29, 2020

Given a binary tree, return all root-to-leaf paths.

Note: A leaf is a node with no children.

Example:

Input:

   1
 /   \
2     3
 \
  5

Output: ["1->2->5", "1->3"]

Explanation: All root-to-leaf paths are: 1->2->5, 1->3

Solution: Recursion

Time complexity: O(n)
Space complexity: O(n) / output can be O(n^2)

C++

// Author: Huahua

class Solution {

public:

vector<string> binaryTreePaths(TreeNode* root) {

vector<string> ans;

string s;

function<void(TreeNode*, int)> preorder = [&](TreeNode* node, int l) {

if (!node) return;

s += (l > 0 ? "->" : "") + to_string(node->val);

if (!node->left && !node->right)

ans.push_back(s);

preorder(node->left, s.size());

preorder(node->right, s.size());

while (s.size() != l) s.pop_back();

};

preorder(root, 0);

return ans;

}

};