Learn from leetCode

2020-01-31 701 words 4 minutes

Contents

Some strategy learning from leetCode

C++ 2d array

Better use a vector(don’t need to consider allocator)

int row = 10;
int col = 10;
int init = 1;
vector<vector <int>> memos(row, vector<int>(col, init)) // init can ignore

int** a = new int*[rowCount];
for(int i = 0; i < rowCount; ++i)
    a[i] = new int[colCount];

delete ...

Traverse a tree

The tree structure:

struct TreeNode {
    int val;
    TreeNode *left;
    TreeNode *right;
    TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};

BFS DFS Pre-order In-order Post-order for a tree

Loop for BFS:

void BFS(TreeNode *root){
    stack<TreeNode *> q;
    q.push(root);
    while(!q.empty()){
        TreeNode *temp = q.front();
        q.pop();
        // do something
        if(temp->left != NULL) q.push(temp -> left);
        if(temp->right != NULL) q.push(temp -> right);
    }
}

Loop for DFS:

void DFS(TreeNode *root){
    stack<TreeNode *> s;
    s.push(root);
    while(!s.empty()){
        TreeNode *temp = s.top();
        s.pop();
        // do something
        if(temp->left != NULL) s.push(temp -> left);
        if(temp->right != NULL) s.push(temp -> right);
    }
}

Pre-order: (DFS is same)

void preOrder(TreeNode *root){
    if(root == NULL) return;
    // root -> val do something
    preOrder(root -> left);
    preOrder(root -> right);
}

In-Order:

void inOrder(TreeNode *root){
    if(root == NULL) return;

    inOrder(root -> left);
    // root -> val do something
    inOrder(root -> right);
}

Pre-Order: (Bottom-to-up)

void postOrder(TreeNode *root){
    if(root == NULL) return;
    postOrder(root -> left);
    postOrder(root -> right);
    // root -> val do something
}

BFS process level by level

void BFS(TreeNode *root){
    stack<TreeNode *> q;
    q.push(root);
    while(!q.empty()){
        int levelSize = q.size();
        // get current level size so that BFS can process level by level
        for(int i=0; i<levelSize; ++i){
            TreeNode *temp = q.front();
            q.pop();
            // do something
            if(temp->left != NULL) q.push(temp -> left);
            if(temp->right != NULL) q.push(temp -> right);
        }
    }
}

heap

Function make_heap can be used to arrange vector into a heap.

It is better to use adopter priority_queue as heap.

#include <queue>

priority_queue<pair<char, int>, vector<pair<char, int>> , less<pair<char,int>> > pq;

Faster io for C++

At this part at beginning.

Relative article: Fast I/O for Competitive Programming - GeeksforGeeks

auto all=[](){
    ios::sync_with_stdio(false); // toggle off synchronization
    cin.tie(nullptr);
    return 0;
}();

Binary search tree(BST) & In-order traversal

After In-order traverse a BST, the result is a sorted sequence.

Also you can use a sorted sequence to rebuild a balance BST.

LeetCode 109 Convert Sorted List to Binary Search Tree

DFS & Backtracking

Example for DFS: LeetCode200

Example for Backtracking: LeetCode46

The different between them is: Backtracking will set the current state back after recursion.

Since they all traverse a matrix, they both need to worry about recursive back. Like DFS goes from left to right and how to protect go back from right left. Of course, sometime it don’t need to be considered.

Word ladder problem

Give a begin word, an end word and a word dictionary.

Word Ladder: leetCode 127

The idea to solve this question is to build a graph (or a State machine) and BFS the graph.

Minimum Genetic Mutation: leetCode 433

Tis question don’t have to build a graph since each char only have 4 options. We can just check the one word mutation is stay in dictionary or not.

Sum O(n^2) -> O(n)

If you want to find all continuous sum in a vector, the naive method is using two for loop:

for (int i = 0; i < size; i++)
{
    int temp_sum = 0;
    for (int j = 0; j < size; j++)
    {
        temp_sum += v[j]
    }
    // temp_sum is sum from i to j. sum(i, j)
}

We can use some tech to make time complexity down to O(n)

The key idea is sum(i, j) = sum(0, j) - sum(0, i-1);

Monotonic Queue

A monotonic queue is a data structure that all elements is strictly increasing or decreasing.

Any new element only entry at the end of the queue and tick out the elements in the queue to make sure the queue is still monotonic.

Example

[5,3,1,2,4]

index	v	Increasing queue	Decreasing queue
1	5	[5]	[5]
2	3	[3]	[5,3]
3	1	[1]	[5,3,1]
4	2	[1,2]	[5,3,2]
5	4	[1,2,4]	[5,4]

Questions

leetcode 581: Shortest Unsorted Continuous Subarray

Convert a number to a binary string (and back) in C++

Reference

Use bitset to easily achieve that.

See my previous post

Question

leetCode 190:Reverse Bits