[Codec] HEVC Study

• Resource
• Documents
• Standard
• ITU H.265 Standard
• Software
• Reference Software: HM
• Encoder: x265
• Viewer/Analyzer
• Intel Video Pro Analyzer
• YUVview
• Video Tool
• GPAC
• ffmpeg
• Reference test stream
• conformance bitstream
• Content
• Profile, Levels, Tiers (introduced in H265)
• Profile
• Main Profile < Main Still Picture Profile < Main 10 Profile
• Main 10 Profile
• 10 bits/sample for HDR, High color gamut, etc.
• Levels
• defines resolution, frame rate, bit rate, buffer capacity (e.g. dpb - decoded picture buffer), ...etc.

• Level	Resolution
  4	HD
  5	4K
  6	8K

• tier
• Main tier/High tier to distinguish the bit rates
• Syntax
• Patent fee

[LeetCode] 118. Pascal's Triangle

• Problem: 118 Pascals's Triangle, 主要為該層元素為上一層的左邊加上一層右邊而得
• Concept:
• 算是滿基本的題目, 稍微將 index 列出來, 可以發現

• A[i][j] = A[i-1][j-1] + A[i-1][j]

• 但還是需要注意 index 從 0開始, 必須習慣一下
• Implementation

• class Solution {
  public:
      vector> generate(int numRows) {
          vector> result;
  
          for (int i = 0; i < numRows; i++) {
              vector row;
              
              row.push_back(1);
              for (int j = 1; j < i; j++) {
                  int val = result[i - 1] [j - 1] + result[i - 1][j];
                  row.push_back(val);
              }
              // no need to add right 1 at first row
              if (i > 0) row.push_back(1);
              
              result.push_back(row);
          }
          return result;
      }
  };
  

[LeetCode] 139. Word Break

• Problem: Word Break
• Concept:
• 可以 recursive 的將一個 word 拆成左半跟右半
• 利用 memorize 的 hash table 來記錄
• 注意終止條件:
• cache 已存在
• 目前的 dictionary 包含完整的 string
• Implementation

• class Solution {
  public:
      bool wordBreak(string s, vector& wordDict) 
      {
          unordered_set quickHash(wordDict.begin(), wordDict.end());
          return wordBreak(s, quickHash);
      }
      
      bool wordBreak(string s, unordered_set quickHash) 
      {
          if (m_mem.count(s)) return m_mem[s];
          if (quickHash.count(s) > 0) {
              m_mem[s] = true;
              return m_mem[s];
          }
  
          for (int i = 1; i < s.length(); i++) {
              if (quickHash.count(s.substr(0, i)) && wordBreak(s.substr(i), quickHash)) {
                  m_mem[s] = true;
                  return m_mem[s];
              }
          }
  
          m_mem[s] = false;
          return m_mem[s];
      }
  
  private:
      unordered_map m_mem;
  };