easy – Page 72 – Huahua’s Tech Road

花花酱 LeetCode 796. Rotate String

By zxi on March 10, 2018

题目大意：给你两个字符串A, B, 问能否通过旋转A得到B。

Problem:

https://leetcode.com/problems/rotate-string/description/

We are given two strings, A and B.

A shift on A consists of taking string A and moving the leftmost character to the rightmost position. For example, if A = 'abcde', then it will be 'bcdea' after one shift on A. Return True if and only if A can become B after some number of shifts on A.

Example 1:

Input: A = 'abcde', B = 'cdeab'

Output: true

Example 2:

Input: A = 'abcde', B = 'abced'

Output: false

Note:

A and B will have length at most 100.

Solution 1: Brute Force

Time complexity: O(n^2)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

bool rotateString(string A, string B) {

if (A.length() != B.length()) return false;

for (int i = 1; i < A.length(); ++i)

if (check(A, B, i)) return true;

return false;

}

private:

bool check(const string& A, const string& B, int offset) {

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

if (A[(i + offset) % A.length()] != B[i])

return false;

return true;

}

};

花花酱 LeetCode 206. Reverse Linked List

By zxi on March 8, 2018

题目大意：反转一个单向链表

Problem:

https://leetcode.com/problems/reverse-linked-list/description/

Reverse a singly linked list.

Solution 1:

Tracking prev / curr / next node

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 9 ms

class Solution {

public:

ListNode* reverseList(ListNode* head) {

ListNode* prev = nullptr;

ListNode* curr = head;

ListNode* next;

while (curr) {

next = curr->next;

curr->next = prev;

prev = curr;

curr = next;

}

return prev;

}

};

Java

// Author: Huahua

// Running time: 0 ms

class Solution {

public ListNode reverseList(ListNode head) {

ListNode prev = null;

ListNode curr = head;

ListNode next;

while (curr != null) {

next = curr.next;

curr.next = prev;

prev = curr;

curr = next;

}

return prev;

}

Python 3

"""

Author: Huahua

Running time: 48 ms

"""

class Solution:

def reverseList(self, head):

prev, curr, nxt = None, head, None;

while curr:

nxt, curr.next = curr.next, prev

prev, curr = curr, nxt

return prev

花花酱 LeetCode 521. Longest Uncommon Subsequence I

By zxi on March 7, 2018

题目大意：求2个字符串最长的不同子序列的长度。

Problem:

https://leetcode.com/problems/longest-uncommon-subsequence-i/description/

Given a group of two strings, you need to find the longest uncommon subsequence of this group of two strings. The longest uncommon subsequence is defined as the longest subsequence of one of these strings and this subsequence should not be any subsequence of the other strings.

A subsequence is a sequence that can be derived from one sequence by deleting some characters without changing the order of the remaining elements. Trivially, any string is a subsequence of itself and an empty string is a subsequence of any string.

The input will be two strings, and the output needs to be the length of the longest uncommon subsequence. If the longest uncommon subsequence doesn’t exist, return -1.

Example 1:

Input: "aba", "cdc"

Output: 3

Explanation: The longest uncommon subsequence is "aba" (or "cdc"),

because "aba" is a subsequence of "aba",

but not a subsequence of any other strings in the group of two strings.

Note:

Both strings’ lengths will not exceed 100.
Only letters from a ~ z will appear in input strings.

Idea:

If two strings are the same, then the longest uncommon sequence does not exist, return -1.

e.g. aaa vs aaa, return -1

Otherwise, the longer string is always a uncommon sequence of the shorter one.

e.g. aaab vs aaa, return 4

Solution 1:

Time complexity: O(n)

Space complexity: O(1)

C++

// Author: Huahua

// Running time: 3 ms

class Solution {

public:

int findLUSlength(string a, string b) {

return a == b ? -1 : max(a.length(), b.length());

}

};

Java

// Author: Huahua

// Running time: 3 ms

class Solution {

public int findLUSlength(String a, String b) {

return a.equals(b) ? -1 : Math.max(a.length(), b.length());

}

Python3

"""

Author: Huahua

Running time: 40 ms (beats 100%)

Website: http://zxi.mytechroad.com/blog/string/leetcode-521-longest-uncommon-subsequence-i/

"""

class Solution:

def findLUSlength(self, a, b):

la, lb = len(a), len(b)

if la != lb: return max(la, lb)

return -1 if a == b else la

花花酱 LeetCode 620. Not Boring Movies

By zxi on March 7, 2018

Problem:

https://leetcode.com/problems/not-boring-movies/description/

X city opened a new cinema, many people would like to go to this cinema. The cinema also gives out a poster indicating the movies’ ratings and descriptions.

Please write a SQL query to output movies with an odd numbered ID and a description that is not ‘boring’. Order the result by rating.

For example, table cinema:

+---------+-----------+--------------+-----------+

+---------+-----------+--------------+-----------+

| 1 | War | great 3D | 8.9 |

| 2 | Science | fiction | 8.5 |

| 3 | irish | boring | 6.2 |

| 4 | Ice song | Fantacy | 8.6 |

| 5 | House card| Interesting| 9.1 |

+---------+-----------+--------------+-----------+

For the example above, the output should be:

+---------+-----------+--------------+-----------+

+---------+-----------+--------------+-----------+

| 5 | House card| Interesting| 9.1 |

| 1 | War | great 3D | 8.9 |

+---------+-----------+--------------+-----------+

MySQL

# Author: Huahua

SELECT * FROM cinema

WHERE id % 2 = 1

AND description != 'boring'

ORDER BY rating DESC

Posts tagged as “easy”

花花酱 LeetCode 234. Palindrome Linked List

花花酱 LeetCode 796. Rotate String

花花酱 LeetCode 206. Reverse Linked List

花花酱 LeetCode 521. Longest Uncommon Subsequence I

花花酱 LeetCode 620. Not Boring Movies