LeetCode设计题 design（共38题）

Posted 2021-02-04 zhangwanying

【146】LRU Cache 

【155】Min Stack 

 

【170】Two Sum III - Data structure design （2018年12月6日，周四）

Design and implement a TwoSum class. It should support the following operations: add and find.

add - Add the number to an internal data structure.
find - Find if there exists any pair of numbers which sum is equal to the value.

Example 1:
add(1); add(3); add(5);
find(4) -> true
find(7) -> false
Example 2:
add(3); add(1); add(2);
find(3) -> true
find(6) -> false

题解：无

 1 class TwoSum {
 2 public:
 3     /** Initialize your data structure here. */
 4     TwoSum() {
 5         
 6     }
 7     
 8     /** Add the number to an internal data structure.. */
 9     void add(int number) {
10         nums.push_back(number);
11         mp[number]++;
12     }
13     
14     /** Find if there exists any pair of numbers which sum is equal to the value. */
15     bool find(int value) {
16         for (auto ele : nums) {
17             int target = value - ele;
18             if (mp.find(target) != mp.end()) {
19                 if (target == ele && mp[target] >= 2 || target != ele && mp[target] >= 1) {
20                     return true;
21                 }
22             }
23         }
24         return false;
25     }
26     vector<int> nums;
27     unordered_map<int, int> mp;
28 };
29 
30 /**
31  * Your TwoSum object will be instantiated and called as such:
32  * TwoSum obj = new TwoSum();
33  * obj.add(number);
34  * bool param_2 = obj.find(value);
35  */

View Code

 

【173】Binary Search Tree Iterator 

【208】Implement Trie (Prefix Tree) （以前 trie 专题做过）

【211】Add and Search Word - Data structure design 

【225】Implement Stack using Queues 

【232】Implement Queue using Stacks 

【244】Shortest Word Distance II 

【251】Flatten 2D Vector 

 

【281】Zigzag Iterator (2019年1月18日，学习什么是iterator)

给了两个一维数组，逐个返回他们的元素。

Example:
Input:
v1 = [1,2]
v2 = [3,4,5,6] 
Output: [1,3,2,4,5,6]
Explanation: By calling next repeatedly until hasNext returns false, the order of elements returned by next should be: [1,3,2,4,5,6].

题解：什么是iterator，就是一个东西它有两个方法：（1）getNext(), (2) hasNext()

本题没有什么好说的，直接搞就行了。

 1 class ZigzagIterator {
 2 public:
 3     ZigzagIterator(vector<int>& v1, vector<int>& v2) {
 4         n1 = v1.size();
 5         n2 = v2.size();
 6         this->v1 = v1, this->v2 = v2;
 7     }
 8     int next() {
 9         if (p1 < n1 && p2 < n2) {
10             int ret = cur == 0 ? v1[p1++] : v2[p2++];
11             cur = 1 - cur;
12             return ret;
13         }
14         if (p1 < n1) {
15             return v1[p1++];
16         } 
17         if (p2 < n2) {
18             return v2[p2++];
19         }
20         return -1;
21     }
22     bool hasNext() {
23         return p1 < n1 || p2 < n2;
24     }
25     int cur = 0;
26     int p1 = 0, p2 = 0, n1 = 0, n2 = 0;
27     vector<int> v1, v2;
28 };
29 /**
30  * Your ZigzagIterator object will be instantiated and called as such:
31  * ZigzagIterator i(v1, v2);
32  * while (i.hasNext()) cout << i.next();
33  */

View Code

 

【284】Peeking Iterator （2019年1月18日，学习什么是iterator）

给了一个 Iterator 的基类，实现 PeekIterator 这个类。它除了有 hashNext() 和 next() 方法之外还有一个方法叫做 peek()，能访问没有访问过的第一个元素，并且不让它pass。

题解：用两个变量保存 是不是有一个 peek 元素，peek元素是什么。每次调用peek方法的时候先查看当前缓存有没有这个元素，如果有的话就返回，如果没有的话，就调用 Iterator::next方法获取当前缓存。 

 1 // Below is the interface for Iterator, which is already defined for you.
 2 // **DO NOT** modify the interface for Iterator.
 3 class Iterator {
 4     struct Data;
 5     Data* data;
 6 public:
 7     Iterator(const vector<int>& nums);
 8     Iterator(const Iterator& iter);
 9     virtual ~Iterator();
10     // Returns the next element in the iteration.
11     int next();
12     // Returns true if the iteration has more elements.
13     bool hasNext() const;
14 };
15 
16 
17 class PeekingIterator : public Iterator {
18 public:
19     PeekingIterator(const vector<int>& nums) : Iterator(nums) {
20         // Initialize any member here.
21         // **DO NOT** save a copy of nums and manipulate it directly.
22         // You should only use the Iterator interface methods.
23         
24     }
25 
26     // Returns the next element in the iteration without advancing the iterator.
27     int peek() {
28         if (hasPeeked) {
29             return peekElement;
30         }
31         peekElement = Iterator::next();
32         hasPeeked = true;
33         return peekElement;
34     }
35 
36     // hasNext() and next() should behave the same as in the Iterator interface.
37     // Override them if needed.
38     int next() {
39         if (hasPeeked) {
40             hasPeeked = false;
41             return peekElement;
42         }
43         return Iterator::next();
44     }
45 
46     bool hasNext() const {
47         if (hasPeeked || Iterator::hasNext()) {
48             return true;
49         }
50         return false;
51     }
52     bool hasPeeked = false; 
53     int peekElement;
54     
55 };

View Code

  

【288】Unique Word Abbreviation 

【295】Find Median from Data Stream 

【297】Serialize and Deserialize Binary Tree 

【341】Flatten Nested List Iterator 

【346】Moving Average from Data Stream 

【348】Design Tic-Tac-Toe 

【353】Design Snake Game 

【355】Design Twitter 

【359】Logger Rate Limiter 

【362】Design Hit Counter 

【379】Design Phone Directory 

【380】Insert Delete GetRandom O(1) 

【381】Insert Delete GetRandom O(1) - Duplicates allowed 

【432】All O`one Data Structure 

【460】LFU Cache 

【588】Design In-Memory File System 

【604】Design Compressed String Iterator 

 

【622】Design Circular Queue (2018年12月12日，算法群，用数组实现队列)

设计用数组实现节省空间的环形队列。

题解：用个size来标记现在队列里面有几个元素，（不要用rear和front的关系来判断，不然一大一小还有环非常容易出错）。

 1 class MyCircularQueue {
 2 public:
 3     /** Initialize your data structure here. Set the size of the queue to be k. */
 4     MyCircularQueue(int k) {
 5         que.resize(k);
 6         capacity = k;
 7         front = rear = 0;
 8         size = 0;
 9     }
10     
11     /** Insert an element into the circular queue. Return true if the operation is successful. */
12     bool enQueue(int value) {
13         if (size == capacity) {return false;}
14         ++size;
15         rear = rear % capacity;
16         que[rear++] = value;
17         return true;
18     }
19     
20     /** Delete an element from the circular queue. Return true if the operation is successful. */
21     bool deQueue() {
22         if (size == 0) {return false;}
23         --size;
24         front = (front + 1) % capacity;
25         return true;
26     }
27     
28     /** Get the front item from the queue. */
29     int Front() {
30         if (size == 0) {return -1;}
31         return que[front];
32     }
33     
34     /** Get the last item from the queue. */
35     int Rear() {
36         if (size == 0) {return -1;}
37         rear = rear % capacity;
38         return rear == 0 ? que[capacity-1] : que[rear-1];
39     }
40     
41     /** Checks whether the circular queue is empty or not. */
42     bool isEmpty() {
43         return size == 0;
44     }
45     
46     /** Checks whether the circular queue is full or not. */
47     bool isFull() {
48         return size == capacity;
49     }
50     int front, rear, size, capacity;
51     vector<int> que;
52 };
53 
54 /**
55  * Your MyCircularQueue object will be instantiated and called as such:
56  * MyCircularQueue* obj = new MyCircularQueue(k);
57  * bool param_1 = obj->enQueue(value);
58  * bool param_2 = obj->deQueue();
59  * int param_3 = obj->Front();
60  * int param_4 = obj->Rear();
61  * bool param_5 = obj->isEmpty();
62  * bool param_6 = obj->isFull();
63  */

View Code

 

【631】Design Excel Sum Formula 

【635】Design Log Storage System 

【641】Design Circular Deque 

【642】Design Search Autocomplete System 

【705】Design HashSet 

【706】Design HashMap 

【707】Design Linked List 

【716】Max Stack