花花酱 LeetCode 753. Cracking the Safe

By zxi on December 31, 2017

题目大意：让你构建一个最短字符串包含所有可能的密码。

There is a box protected by a password. The password is n digits, where each letter can be one of the first kdigits 0, 1, ..., k-1.

You can keep inputting the password, the password will automatically be matched against the last n digits entered.

For example, assuming the password is "345", I can open it when I type "012345", but I enter a total of 6 digits.

Please return any string of minimum length that is guaranteed to open the box after the entire string is inputted.

Example 1:

Input: n = 1, k = 2

Output: "01"

Note: "10" will be accepted too.

Example 2:

Input: n = 2, k = 2

Output: "00110"

Note: "01100", "10011", "11001" will be accepted too.

Note:

n will be in the range [1, 4].
k will be in the range [1, 10].
k^n will be at most 4096.

Idea: Search

Solution 1: DFS w/ backtracking

C ++

// Author: Huahua

// Runtime: 9 ms

class Solution {

public:

string crackSafe(int n, int k) {

const int total_len = pow(k, n) + n - 1;

string ans(n, '0');

unordered_set<string> visited{ans};

if (dfs(ans, total_len, n, k, visited))

return ans;

return "";

}

private:

bool dfs(string& ans, const int total_len, const int n, const int k, unordered_set<string>& visited) {

if (ans.length() == total_len)

return true;

string node = ans.substr(ans.length() - n + 1, n - 1);

for (char c = '0'; c < '0' + k; ++c) {

node.push_back(c);

if (!visited.count(node)) {

ans.push_back(c);

visited.insert(node);

if (dfs(ans, total_len, n, k, visited)) return true;

visited.erase(node);

ans.pop_back();

}

node.pop_back();

}

return false;

}

};

Solution 2: DFS w/o backtracking

C++

// Author: Huahua

// Running time: 13 ms

class Solution {

public:

string crackSafe(int n, int k) {

const int total_len = pow(k, n) + n - 1;

string node(n - 1, '0');

string ans;

unordered_set<string> visited;

dfs(node, k, visited, ans);

return ans + node;

}

private:

void dfs(const string& node, const int k, unordered_set<string>& visited, string& ans) {

for (char c = '0'; c < '0' + k; ++c) {

string next = node + c;

if (visited.count(next)) continue;

visited.insert(next);

dfs(next.substr(1), k, visited, ans);

ans.push_back(c);

}

};

花花酱 LeetCode 476. Number Complement

By zxi on December 29, 2017

题目大意：给你一个正整数，输出和它互补的数（翻转所有的bits）。

Problem:

Given a positive integer, output its complement number. The complement strategy is to flip the bits of its binary representation.

Note:

The given integer is guaranteed to fit within the range of a 32-bit signed integer.
You could assume no leading zero bit in the integer’s binary representation.

Example 1:

Input: 5

Output: 2

Explanation: The binary representation of 5 is 101 (no leading zero bits), and its complement is 010. So you need to output 2.

Example 2:

Input: 1

Output: 0

Explanation: The binary representation of 1 is 1 (no leading zero bits), and its complement is 0. So you need to output 0.

Idea:

Bit

Solution:

C++

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

int findComplement(int num) {

int mask = ~0;

while (num & mask) mask <<= 1;

return ~num ^ mask;

}

};

花花酱 LeetCode 367. Valid Perfect Square

By zxi on December 29, 2017

题目大意：判断一个数是否是平方数。不能使用开根号函数。

Given a positive integer num, write a function which returns True if num is a perfect square else False.

Note: Do not use any built-in library function such as sqrt.

Example 1:

1 2	Input: 16 Returns: True

Example 2:

1 2	Input: 14 Returns: False

Idea:

Binary search

Solution:

C++

Time complexity: O(log(num))

Space complexity: O(1)

// Author: Huahua

// Runtime: 0 ms

class Solution {

public:

bool isPerfectSquare(int num) {

long l = 1;

long r = num + 1;

while (l < r) {

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

long t = m * m;

if (t == num)

return true;

else if (t > num)

r = m;

else

l = m + 1;

}

return false;

}

};

花花酱 LeetCode 653. Two Sum IV – Input is a BST

By zxi on December 29, 2017

题目大意：给你一棵二叉搜索树，返回树中是否存在两个节点的和等于给定的目标值。

Problem:

Given a Binary Search Tree and a target number, return true if there exist two elements in the BST such that their sum is equal to the given target.

Example 1:

Input:

/ \

3 6

/ \ \

2 4 7

Target = 9

Output: True

Example 2:

Input:

/ \

3 6

/ \ \

2 4 7

Target = 28

Output: False

Solution:

C++

// Author: Huahua

// Running time: 36 ms

class Solution {

public:

bool findTarget(TreeNode* root, int k) {

vector<int> nums;

inorder(root, nums);

int l = 0;

int r = nums.size() - 1;

while (l < r) {

int sum = nums[l] + nums[r];

if (sum == k) return true;

else if (sum < k)

++l;

else

--r;

}

return false;

}

private:

void inorder(TreeNode* root, vector<int>& nums) {

if (root == nullptr) return;

inorder(root->left, nums);

nums.push_back(root->val);

inorder(root->right, nums);

}

};

花花酱 LeetCode 748. Largest Number At Least Twice of Others

By zxi on December 29, 2017

题目大意：给你一个数组，问你其中最大的数是不是比剩下的数大2倍以上，返回这样的数的索引。如果不存在，返回-1.

Problem:

In a given integer array nums, there is always exactly one largest element.

Find whether the largest element in the array is at least twice as much as every other number in the array.

If it is, return the index of the largest element, otherwise return -1.

Example 1:

Input: nums = [3, 6, 1, 0]

Output: 1

Explanation: 6 is the largest integer, and for every other number in the array x,

6 is more than twice as big as x. The index of value 6 is 1, so we return 1.

Example 2:

Input: nums = [1, 2, 3, 4]

Output: -1

Explanation: 4 isn't at least as big as twice the value of 3, so we return -1.

Note:

nums will have a length in the range [1, 50].
Every nums[i] will be an integer in the range [0, 99].

Solution:

C++

// Author: Huahua

// Running time: 6 ms

class Solution {

public:

int dominantIndex(vector<int>& nums) {

auto it1 = max_element(nums.begin(), nums.end());

const int max1 = *it1;

*it1 = -1;

const int max2 = *max_element(nums.begin(), nums.end());

return max1 >= max2 * 2 ? distance(nums.begin(), it1) : -1;

}

};

Huahua's Tech Road

花花酱 LeetCode 753. Cracking the Safe

Idea: Search

Solution 1: DFS w/ backtracking

C ++

Solution 2: DFS w/o backtracking

C++

花花酱 LeetCode 476. Number Complement

花花酱 LeetCode 367. Valid Perfect Square

花花酱 LeetCode 653. Two Sum IV – Input is a BST

花花酱 LeetCode 748. Largest Number At Least Twice of Others