花花酱 LeetCode 557 Reverse Words in a String III

By zxi on March 2, 2018

题目大意：独立反转字符串中的每个单词。

Given a string, you need to reverse the order of characters in each word within a sentence while still preserving whitespace and initial word order.

Example 1:

1 2	Input: "Let's take LeetCode contest" Output: "s'teL ekat edoCteeL tsetnoc"

Note: In the string, each word is separated by single space and there will not be any extra space in the string.

Idea: Brute Force

C++

// Author: Huahua

// Running time: 23 ms

class Solution {

public:

string reverseWords(string s) {

int index = 0;

for (int i = 0; i <= s.length(); ++i) {

if (i == s.length() || s[i] == ' ') {

std::reverse(s.begin() + index, s.begin() + i);

index = i + 1;

}

return s;

}

};

Python3

"""

Author: Huahua

Running time: 44 ms (beats 100%)

"""

class Solution:

def reverseWords(self, s):

return ' '.join(map(lambda x: x[::-1], s.split(' ')))

花花酱 LeetCode 508. Most Frequent Subtree Sum

By zxi on March 2, 2018

Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1
Input:

/ \

2 -3

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2
Input:

/ \

2 -5

return [2], since 2 happens twice, however -5 only occur once.

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

// Author: Huahua

// Running time: 18 ms

class Solution {

public:

vector<int> findFrequentTreeSum(TreeNode* root) {

unordered_map<int, int> freqs;

int max_freq = -1;

vector<int> ans;

(void)treeSum(root, freqs, max_freq, ans);

return ans;

}

private:

int treeSum(TreeNode* root, unordered_map<int, int>& freqs, int& max_freq, vector<int>& ans) {

if (!root) return 0;

int sum = root->val +

treeSum(root->left, freqs, max_freq, ans) +

treeSum(root->right, freqs, max_freq, ans);

int freq = ++freqs[sum];

if (freq > max_freq) {

max_freq = freq;

ans.clear();

}

if (freq == max_freq)

ans.push_back(sum);

return sum;

}

};

Python

"""

Author: Huahua

Running time: 72 ms (beats 100%)

"""

class Solution:

def findFrequentTreeSum(self, root):

if not root: return []

def treeSum(root):

if not root: return 0

s = root.val + treeSum(root.left) + treeSum(root.right)

f[s] += 1

return s

f = collections.Counter()

treeSum(root)

max_freq = max(f.values())

return [s for s in f.keys() if f[s] == max_freq]

花花酱 LeetCode 518. Coin Change 2

By zxi on March 2, 2018

题目大意：给你一些硬币的面值，问使用这些硬币（无限多块）能够组成amount的方法有多少种。

You are given coins of different denominations and a total amount of money. Write a function to compute the number of combinations that make up that amount. You may assume that you have infinite number of each kind of coin.

Note: You can assume that

0 <= amount <= 5000
1 <= coin <= 5000
the number of coins is less than 500
the answer is guaranteed to fit into signed 32-bit integer

Example 1:

Input: amount = 5, coins = [1, 2, 5]

Output: 4

Explanation: there are four ways to make up the amount:

5=5

5=2+2+1

5=2+1+1+1

5=1+1+1+1+1

Example 2:

Input: amount = 3, coins = [2]

Output: 0

Explanation: the amount of 3 cannot be made up just with coins of 2.

Example 3:

1 2	Input: amount = 10, coins = [10] Output: 1

Idea: DP

Transition 1:

Let us use dp[i][j] to denote the number of ways to sum up to amount j using first i kind of coins.

dp[i][j] = dp[i – 1][j – coin] + dp[i – 1][j – 2* coin] + …

Time complexity: O(n*amount^2) TLE

Space complexity: O(n*amount) -> O(amount)

Transition 2:

Let us use dp[i] to denote the number of ways to sum up to amount i.

dp[i + coin] += dp[i]

Time complexity: O(n*amount)

Space complexity: O(amount)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int change(int amount, vector<int>& coins) {

vector<int> dp(amount + 1, 0);

dp[0] = 1;

for (const int coin : coins)

for (int i = 0; i <= amount - coin; ++i)

dp[i + coin] += dp[i];

return dp[amount];

}

};

Java

// Author: Huahua

// Running time: 7 ms

class Solution {

public int change(int amount, int[] coins) {

int[] dp = new int[amount + 1];

dp[0] = 1;

for (int coin : coins)

for (int i = 0; i <= amount - coin; ++i)

dp[i + coin] += dp[i];

return dp[amount];

}

Python

"""

Author: Huahua

Running time: 107 ms (beats 100%)

"""

class Solution(object):

def change(self, amount, coins):

dp = [1] + [0] * (amount);

for coin in coins:

for i in xrange(amount - coin + 1):

if dp[i]:

dp[i + coin] += dp[i]

return dp[amount]

Related Problems:

花花酱 LeetCode 322. Coin Change

花花酱 LeetCode 468. Validate IP Address

By zxi on March 1, 2018

题目大意：给你一个字符串，让你判断是否是合法的ipv4或者ipv6地址，都不合法返回Neighter.

Write a function to check whether an input string is a valid IPv4 address or IPv6 address or neither.

IPv4 addresses are canonically represented in dot-decimal notation, which consists of four decimal numbers, each ranging from 0 to 255, separated by dots (“.”), e.g.,172.16.254.1;

Besides, leading zeros in the IPv4 is invalid. For example, the address 172.16.254.01 is invalid.

IPv6 addresses are represented as eight groups of four hexadecimal digits, each group representing 16 bits. The groups are separated by colons (“:”). For example, the address 2001:0db8:85a3:0000:0000:8a2e:0370:7334 is a valid one. Also, we could omit some leading zeros among four hexadecimal digits and some low-case characters in the address to upper-case ones, so 2001:db8:85a3:0:0:8A2E:0370:7334 is also a valid IPv6 address(Omit leading zeros and using upper cases).

However, we don’t replace a consecutive group of zero value with a single empty group using two consecutive colons (::) to pursue simplicity. For example, 2001:0db8:85a3::8A2E:0370:7334 is an invalid IPv6 address.

Besides, extra leading zeros in the IPv6 is also invalid. For example, the address 02001:0db8:85a3:0000:0000:8a2e:0370:7334 is invalid.

Note: You may assume there is no extra space or special characters in the input string.

Example 1:

Input: "172.16.254.1"

Output: "IPv4"

Explanation: This is a valid IPv4 address, return "IPv4".

Example 2:

Input: "2001:0db8:85a3:0:0:8A2E:0370:7334"

Output: "IPv6"

Explanation: This is a valid IPv6 address, return "IPv6".

Example 3:

Input: "256.256.256.256"

Output: "Neither"

Explanation: This is neither a IPv4 address nor a IPv6 address.

Solution 1: String / Brute force

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

bool isV4Num(string n) {

if (n.length() > 3 || n.empty()) return false;

std::reverse(n.begin(), n.end());

int a = 0;

int f = 1;

for (char c : n) {

if (!isdigit(c)) return false;

if (c == '0' && f > 1) return false;

a += (c - '0') * f;

f *= 10;

}

return a >= 0 && a <= 255;

}

bool isV6Num(string n) {

if (n.length() > 4 || n.empty()) return false;

std::reverse(n.begin(), n.end());

int a = 0;

int f = 1;

for (char c : n) {

if (isdigit(c))

a += (c - '0') * f;

else if (c >= 'a' && c <= 'f' || c >= 'A' && c <= 'Z')

a += (tolower(c) - 'a' + 10) * f;

else

return false;

f *= 16;

}

return a >= 0 && a < (1 << 16);

}

string validIPAddress(string IP) {

bool isIPv4 = true;

bool isIPv6 = true;

string cur;

int seg = 0;

for (char c: IP) {

if (c == '.') { ++seg; isIPv6 = false; isIPv4 &= isV4Num(cur); cur.clear(); }

else if (c == ':') { ++seg; isIPv4 = false; isIPv6 &= isV6Num(cur); cur.clear(); }

else cur += c;

}

isIPv4 &= isV4Num(cur) && seg == 3;

isIPv6 &= isV6Num(cur) && seg == 7;

return isIPv4 ? "IPv4" : (isIPv6 ? "IPv6" : "Neither");

}

};

花花酱 LeetCode 790. Domino and Tromino Tiling

By zxi on February 24, 2018

题目大意：有两种不同形状的骨牌(1×2长条形，L型）无限多块。给你一个2xN的板子，问一共有多少不同的方式可以完全覆盖。

We have two types of tiles: a 2×1 domino shape, and an “L” tromino shape. These shapes may be rotated.

XX <- domino

XX <- "L" tromino

Given N, how many ways are there to tile a 2 x N board? Return your answer modulo 10^9 + 7.

(In a tiling, every square must be covered by a tile. Two tilings are different if and only if there are two 4-directionally adjacent cells on the board such that exactly one of the tilings has both squares occupied by a tile.)

Example:

Input: 3

Output: 5

Explanation:

The five different ways are listed below, different letters indicates different tiles:

XYZ XXZ XYY XXY XYY

XYZ YYZ XZZ XYY XXY

Idea: DP

dp[i][0]: ways to cover i cols, both rows of i-th col are covered
dp[i][1]: ways to cover i cols, only top row of i-th col is covered
dp[i][2]: ways to cover i cols, only bottom row of i-th col is covered

Solution 1: DP

Time complexity: O(N)

Space complexity: O(N)

C++

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int numTilings(int N) {

constexpr int kMod = 1000000007;

vector<vector<long>> dp(N + 1, vector<long>(3, 0));

dp[0][0] = dp[1][0] = 1;

for (int i = 2; i <= N; ++i) {

dp[i][0] = (dp[i - 1][0] + dp[i - 2][0] + dp[i - 1][1] + dp[i - 1][2]) % kMod;

dp[i][1] = (dp[i - 2][0] + dp[i - 1][2]) % kMod;

dp[i][2] = (dp[i - 2][0] + dp[i - 1][1]) % kMod;

}

return dp[N][0];

}

};

C++ V2

Since dp[i][1] always equals to dp[i][2], we can simplify a bit.

// Author: Huahua

// Running time: 4 ms

class Solution {

public:

int numTilings(int N) {

constexpr int kMod = 1000000007;

vector<vector<long>> dp(N + 1, vector<long>(2, 0));

dp[0][0] = dp[1][0] = 1;

for (int i = 2; i <= N; ++i) {

dp[i][0] = (dp[i - 1][0] + dp[i - 2][0] + 2 * dp[i - 1][1]) % kMod;

dp[i][1] = (dp[i - 2][0] + dp[i - 1][1]) % kMod;

}

return dp[N][0];

}

};

Solution 2: DP

Another way to think about this problem

define: dp[i] ways to completely covert the i*2 board.

dp[0] = 1 # {}

dp[1] = 1 # {|}

dp[2] = 2 # {||, =}

dp[3] = 5 # {|||, |=, =|, ⌊⌉, ⌈⌋} = dp[2] ⊗ {|} + dp[1] ⊗ {=} + dp[0] ⊗ {⌊⌉, ⌈⌋}

dp[4] = 11 # dp[3] ⊗ {|} + dp[2] ⊗ {=} + dp[1] ⊗ {⌊⌉, ⌈⌋} + dp[0] ⊗ {⌊¯⌋,⌈_⌉}

dp[5] = 24 # dp[4] ⊗ {|} + dp[3] ⊗ {=} + 2*(dp[2] + dp[1] + dp[0])

...

dp[n] = dp[n-1] + dp[n-2] + 2*(dp[n-3] + ... + dp[0])

= dp[n-1] + dp[n-3] + [dp[n-2] + dp[n-3] + 2*(dp[n-4] + ... + dp[0])]

= dp[n-1] + dp[n-3] + dp[n-1]

= 2*dp[n-1] + dp[n-3]

C++

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

int numTilings(int N) {

constexpr int kMod = 1000000007;

vector<long> dp(N + 1, 1);

dp[2] = 2;

for (int i = 3; i <= N; ++i)

dp[i] = (dp[i - 3] + dp[i - 1] * 2) % kMod;

return dp[N];

}

};

Huahua's Tech Road

花花酱 LeetCode 557 Reverse Words in a String III

花花酱 LeetCode 508. Most Frequent Subtree Sum

花花酱 LeetCode 518. Coin Change 2

花花酱 LeetCode 468. Validate IP Address

花花酱 LeetCode 790. Domino and Tromino Tiling

Idea: DP

Solution 1: DP

C++

C++ V2

Solution 2: DP

C++