花花酱 LeetCode 572. Subtree of Another Tree

By zxi on March 19, 2018

Problem

题目大意：判断一棵树是不是另外一棵树的子树。

https://leetcode.com/problems/subtree-of-another-tree/description/

Given two non-empty binary trees s and t, check whether tree t has exactly the same structure and node values with a subtree of s. A subtree of s is a tree consists of a node in s and all of this node’s descendants. The tree s could also be considered as a subtree of itself.

Example 1:
Given tree s:

/ \

4 5

/ \

1 2

Given tree t:

/ \

1 2

Return true, because t has the same structure and node values with a subtree of s.

Example 2:
Given tree s:

/ \

4 5

/ \

1 2

Given tree t:

/ \

1 2

Return false.

Solution: Recursion

Time complexity: O(max(n, m))

Space complexity: O(max(n, m))

C++

// Author: Huahua

// Running time: 29 ms

class Solution {

public:

bool isSubtree(TreeNode* s, TreeNode* t) {

if (t == nullptr) return true;

if (s == nullptr) return false;

if (isSameTree(s, t)) return true;

return isSubtree(s->left, t) || isSubtree(s->right, t);

}

private:

bool isSameTree(TreeNode* s, TreeNode* t) {

if (s == nullptr && t == nullptr) return true;

if (s == nullptr || t == nullptr) return false;

return (s->val == t->val)

&& isSameTree(s->left, t->left)

&& isSameTree(s->right, t->right);

}

};

Problem

The Hamming distance between two integers is the number of positions at which the corresponding bits are different.

Given two integers x and y, calculate the Hamming distance.

Note:
0 ≤ x, y < 2³¹.

Example:

Input: x = 1, y = 4

Output: 2

Explanation:
1   (0 0 0 1)
4   (0 1 0 0)
       ↑   ↑

The above arrows point to positions where the corresponding bits are different.

Solution: Bit Operation

Time complexity: O(logn)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 6 ms

class Solution {

public:

int hammingDistance(int x, int y) {

int ans = 0;

int t = x^y;

while (t) {

ans += t & 1;

t >>=1;

}

return ans;

}

};

// Author: Huahua

// Running time: 9 ms

class Solution {

public:

int hammingDistance(int x, int y) {

return __builtin_popcount(x ^ y);

}

};

Problem

In the following, every capital letter represents some hexadecimal digit from 0 to f.

The red-green-blue color "#AABBCC" can be written as "#ABC" in shorthand. For example, "#15c" is shorthand for the color "#1155cc".

Now, say the similarity between two colors "#ABCDEF" and "#UVWXYZ" is -(AB - UV)^2 - (CD - WX)^2 - (EF - YZ)^2.

Given the color "#ABCDEF", return a 7 character color that is most similar to #ABCDEF, and has a shorthand (that is, it can be represented as some "#XYZ"

Example 1:
Input: color = "#09f166"
Output: "#11ee66"
Explanation:  
The similarity is -(0x09 - 0x11)^2 -(0xf1 - 0xee)^2 - (0x66 - 0x66)^2 = -64 -9 -0 = -73.
This is the highest among any shorthand color.

Note:

color is a string of length 7.
color is a valid RGB color: for i > 0, color[i] is a hexadecimal digit from 0 to f
Any answer which has the same (highest) similarity as the best answer will be accepted.
All inputs and outputs should use lowercase letters, and the output is 7 characters.

Solution: Brute Force

R, G, B are independent, find the closest color for each channel separately.

Time complexity: O(3 * 16)

Space complexity: O(1)

// Author: Huahua

// Running time: 7 ms

class Solution {

public:

string similarRGB(string color) {

const string hex{"0123456789abcdef"};

vector<int> rgb(3, 0);

for (int i = 0; i < 3; ++i)

rgb[i] = hex.find(color[2 * i + 1]) * 16 + hex.find(color[2 * i + 2]);

string ans(7, '#');

for (int i = 0; i < 3; ++i) {

int best = INT_MAX;

for (int j = 0; j < 16; ++j) {

int diff = abs(j * 16 + j - rgb[i]);

if (diff >= best) continue;

best = diff;

ans[2 * i + 1] = ans[2 * i + 2] = hex[j];

}

return ans;

}

};

花花酱 LeetCode 415. Add Strings

By zxi on March 17, 2018

Problem

Given two non-negative integers num1 and num2 represented as string, return the sum of num1 and num2.

Note:

The length of both num1 and num2 is < 5100.
Both num1 and num2 contains only digits 0-9.
Both num1 and num2 does not contain any leading zero.
You must not use any built-in BigInteger library or convert the inputs to integer directly.

Solution: Brute Force

Time complexity: O(n)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 13 ms

class Solution {

public:

string addStrings(string num1, string num2) {

if (num1.length() < num2.length()) swap(num1, num2);

num1 = "0" + num1;

int l1 = num1.length();

int l2 = num2.length();

for (int i = 0; i < l1 - 1; ++i) {

if (i < l2) num1[l1 - i - 1] += (num2[l2 - i - 1] - '0');

if (num1[l1 - i - 1] > '9') {

num1[l1 - i - 1] -= 10;

++num1[l1 - i - 2];

}

return (num1[0] != '0') ? num1 : num1.substr(1);

}

};

花花酱 LeetCode 414. Third Maximum Number

By zxi on March 17, 2018

Problem

https://leetcode.com/problems/third-maximum-number/description/

Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).

Example 1:

Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.

Example 2:

Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.

Example 3:

Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.

Solution: Set

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 9 ms

class Solution {

public:

int thirdMax(vector<int>& nums) {

set<int> s;

for (int num : nums) {

s.insert(num);

if (s.size() > 3) s.erase(s.begin());

}

return s.size() != 3 ? *s.rbegin() : *s.begin();

}

};

Python3

"""

Author: Huahua

Running time: 52 ms

"""

class Solution:

def thirdMax(self, nums):

s = set()

for num in nums:

s.add(num)

if len(s) > 3: s.remove(min(s))

return min(s) if len(s) == 3 else max(s)

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花花酱 LeetCode 572. Subtree of Another Tree

Problem

Solution: Recursion

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花花酱 LeetCode 414. Third Maximum Number

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