Skip to main content

Recursion

Recursion is one of the most important concepts in algorithm. Once you understand it, it is quite fair to say that you may have mastered almost "half" of the algorithm.

Recursion is based one an obvious fact: computers can do repetitive operations in a much faster way than human beings. In practice, to avoid infinite loop, the terminating condition has to be clear. Thus, the size of the problem to be solved has to become smaller (Neither the same Nor larger) for every recursive call, which eventually converges to the base case. 

The base cases usually have very small sizes which can return the results trivially.

To sum up, some key points:

a). base cases with trivial returns;

b). the same question but a shrinking size.


Question List


Upper Layer

Comments

Post a Comment

Popular posts from this blog

Sweep Line

Sweep (or scanning) line algorithm is very efficient for some specific questions involving discrete intervals. The intervals could be the lasting time of events, or the width of a building or an abstract square, etc. In the scanning line algorithm, we usually need to distinguish the start and the end of an interval. After the labeling of the starts and ends, we can sort them together based on the values of the starts and ends. Thus, if there are N intervals in total, we will have 2*N data points (since each interval will contribute 2). The sorting becomes the most time-consuming step, which is O(2N*log(2N) ~ O(N*logN). After the sorting, we usually can run a linear sweep for all the data points. If the data point is labeled as a starting point, it means a new interval is in the processing; when an ending time is reached, it means one of the interval has ended. In such direct way, we can easily figure out how many intervals are in the processes. Other related information can also be obt...
Post a Comment
Read more

Bit Manipulation - Example

  Leetcode 136 Single Number Given a non-empty array of integers nums, every element appears twice except for one. Find that single one. You must implement a solution with a linear runtime complexity and use only constant extra space. Constraints: 1 <= nums.length <= 3 * 10^4 -3 * 10^4 <= nums[i] <= 3 * 10^4 Each element in the array appears twice except for one element which appears only once. Analysis: If there is no space limitation, this question can be solved by counting easily. But counting requires additional space. Here we can use xor (^) operation based on some interesting observations:  A^A = 0, here A is any number A^0 = A, here A is any number Since all the number appears twice except one, then all the number appear even numbers will be cancelled out, and only the number appears one time is left, which is what we want. See the code below: class Solution { public: int singleNumber(vector<int>& nums) { int res = 0; for(auto ...
Post a Comment
Read more

Algorithm Advance Outline

Who wants to read these notes? 1. the one who wants to learn algorithm; 2. the one who has a planned interview in a short amount of time, such as one month later; 3. purely out of curiosity; 4. all the rest. The primary purpose for these posts is to help anyone who wants to learn algorithm in a short amount of time and be ready for coding interviews with tech companies, such as, Amazon, Facebook, Microsoft, etc. Before you start, you are expected to already know:  1. the fundamentals of at least one programming language; 2. the fundamentals of data structures. If you do not have these basics, it is better to "google and read" some intro docs, which should NOT take large amount of time. Of course, you can always learn whenever see a "unknown" data structure or line in the codes. Remember, "google and read" is always one of keys to learning. Common algorithms: 1. Recursion 2. Binary search 3. Dynamic programming 4. Breadth-first search 5. Depth-first search...
Post a Comment
Read more