Posts tagged as “easy”

花花酱 LeetCode 225. Implement Stack using Queues

By zxi on March 13, 2018

题目大意：用队列来实现栈。

Problem:

https://leetcode.com/problems/implement-stack-using-queues/description/

Implement the following operations of a stack using queues.

push(x) — Push element x onto stack.
pop() — Removes the element on top of the stack.
top() — Get the top element.
empty() — Return whether the stack is empty.

Notes:

You must use only standard operations of a queue — which means only push to back, peek/pop from front, size, and is empty operations are valid.
Depending on your language, queue may not be supported natively. You may simulate a queue by using a list or deque (double-ended queue), as long as you use only standard operations of a queue.
You may assume that all operations are valid (for example, no pop or top operations will be called on an empty stack).

Idea:

Using a single queue, for every push, shift the queue (n – 1) times such that the last element becomes the first element in the queue.

e.g.

push(1): q: [1]

push(2): q: [1, 2] -> [2, 1]

push(3): q: [2, 1, 3] -> [1, 3, 2] -> [3, 2, 1]

push(4): q: [3, 2, 1, 4] -> [2, 1, 4, 3] -> [1, 4, 3, 2] -> [4, 3, 2, 1]

Solution:

Time complexity:

Push: O(n)

Pop/top/empty: O(1)

Space complexity: O(n)

C++

// Author: Huahua

// Running time: 3 ms

class MyStack {

public:

/** Initialize your data structure here. */

MyStack() {}

/** Push element x onto stack. */

void push(int x) {

q_.push(x);

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

q_.push(q_.front());

q_.pop();

}

/** Removes the element on top of the stack and returns that element. */

int pop() {

int top = q_.front();

q_.pop();

return top;

}

/** Get the top element. */

int top() {

return q_.front();

}

/** Returns whether the stack is empty. */

bool empty() {

return q_.empty();

}

private:

queue<int> q_;

};

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

Posts tagged as “easy”

花花酱 LeetCode 225. Implement Stack using Queues

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