Wednesday, May 9, 2018

LeetCode 699 - Falling Squares


https://leetcode.com/problems/falling-squares/solution/
On an infinite number line (x-axis), we drop given squares in the order they are given.
The i-th square dropped (positions[i] = (left, side_length)) is a square with the left-most point being positions[i][0]and sidelength positions[i][1].
The square is dropped with the bottom edge parallel to the number line, and from a higher height than all currently landed squares. We wait for each square to stick before dropping the next.
The squares are infinitely sticky on their bottom edge, and will remain fixed to any positive length surface they touch (either the number line or another square). Squares dropped adjacent to each other will not stick together prematurely.

Return a list ans of heights. Each height ans[i] represents the current highest height of any square we have dropped, after dropping squares represented by positions[0], positions[1], ..., positions[i].
Example 1:
Input: [[1, 2], [2, 3], [6, 1]]
Output: [2, 5, 5]
Explanation:
After the first drop of positions[0] = [1, 2]: _aa _aa ------- The maximum height of any square is 2.
After the second drop of positions[1] = [2, 3]: __aaa __aaa __aaa _aa__ _aa__ -------------- The maximum height of any square is 5. The larger square stays on top of the smaller square despite where its center of gravity is, because squares are infinitely sticky on their bottom edge.
After the third drop of positions[1] = [6, 1]: __aaa __aaa __aaa _aa _aa___a -------------- The maximum height of any square is still 5. Thus, we return an answer of [2, 5, 5].

Example 2:
Input: [[100, 100], [200, 100]]
Output: [100, 100]
Explanation: Adjacent squares don't get stuck prematurely - only their bottom edge can stick to surfaces.
Note:

  • 1 <= positions.length <= 1000.
  • 1 <= positions[i][0] <= 10^8.
  • 1 <= positions[i][1] <= 10^6.

  • X. TreeMap
    https://leetcode.com/problems/falling-squares/discuss/112678/Treemap-solution-and-Segment-tree-(Java)-solution-with-lazy-propagation-and-coordinates-compression
    https://www.geeksforgeeks.org/lazy-propagation-in-segment-tree/
    TreeMap Solution: The basic idea here is pretty simple, for each square i, I will find all the maximum height from previously dropped squares range from floorKey(i_start) (inclusive) to end (exclusive), then I will update the height and delete all the old heights.
        public List<Integer> fallingSquares(int[][] positions) {
            List<Integer> res = new ArrayList<>();
            TreeMap<Integer, Integer> startHeight = new TreeMap<>();
            startHeight.put(0, 0); 
            int max = 0;
            for (int[] pos : positions) {
                int start = pos[0], end = start + pos[1];
                Integer from = startHeight.floorKey(start);
                int height = startHeight.subMap(from, end).values().stream().max(Integer::compare).get() + pos[1];
                max = Math.max(height, max);
                res.add(max);
                // remove interval within [start, end)
                int lastHeight = startHeight.floorEntry(end).getValue();
                startHeight.put(start, height);
                startHeight.put(end, lastHeight);
                startHeight.keySet().removeAll(new HashSet<>(startHeight.subMap(start, false, end, false).keySet()));
            }
            return res;
        }
    https://leetcode.com/problems/falling-squares/discuss/108775/Easy-Understood-TreeMap-Solution
    The squares divide the number line into many segments with different heights. Therefore we can use a TreeMap to store the number line. The key is the starting point of each segment and the value is the height of the segment. For every new falling square (s, l), we update those segments between s and s + l.
        public List<Integer> fallingSquares(int[][] positions) {
            List<Integer> list = new ArrayList<>();
            TreeMap<Integer, Integer> map = new TreeMap<>();
    
            // at first, there is only one segment starting from 0 with height 0
            map.put(0, 0);
            
            // The global max height is 0
            int max = 0;
    
            for(int[] position : positions) {
    
                // the new segment 
                int start = position[0], end = start + position[1];
    
                // find the height among this range
                Integer key = map.floorKey(start);
                int h = map.get(key);
                key = map.higherKey(key);
                while(key != null && key < end) {
                    h = Math.max(h, map.get(key));
                    key = map.higherKey(key);
                }
                h += position[1];
    
                // update global max height
                max = Math.max(max, h);
                list.add(max);
    
                // update new segment and delete previous segments among the range
                int tail = map.floorEntry(end).getValue();
                map.put(start, h);
                map.put(end, tail);
                key = map.higherKey(start);
                while(key != null && key < end) {
                    map.remove(key);
                    key = map.higherKey(key);
                }
            }
            return list;
        }
    https://leetcode.com/problems/falling-squares/discuss/108769/C++-O(nlogn)
    Similar to skyline concept, going from left to right the path is decomposed to consecutive segments, and each segment has a height. Each time we drop a new square, then update the level map by erasing & creating some new segments with possibly new height. There are at most 2n segments that are created / removed throughout the process, and the time complexity for each add/remove operation is O(log(n)).
    X. https://leetcode.com/problems/falling-squares/discuss/108766/Easy-Understood-O(n2)-Solution-with-explanation
    The idea is quite simple, we use intervals to represent the square. the initial height we set to the square cur is pos[1]. That means we assume that all the square will fall down to the land. we iterate the previous squares, check is there any square i beneath my cur square. If we found that we have squares i intersect with us, which means my current square will go above to that square i. My target is to find the highest square and put square cur onto square i, and set the height of the square cur as
    cur.height = cur.height + previousMaxHeight;
    

        private class Interval {
            int start, end, height;
            public Interval(int start, int end, int height) {
                this.start = start;
                this.end = end;
                this.height = height;
            }
        }
        public List<Integer> fallingSquares(int[][] positions) {
            List<Interval> intervals = new ArrayList<>();
            List<Integer> res = new ArrayList<>();
            int h = 0;
            for (int[] pos : positions) {
                Interval cur = new Interval(pos[0], pos[0] + pos[1] - 1, pos[1]);
                h = Math.max(h, getHeight(intervals, cur));
                res.add(h);
            }
            return res;
        }
        private int getHeight(List<Interval> intervals, Interval cur) {
            int preMaxHeight = 0;
            for (Interval i : intervals) {
                // Interval i does not intersect with cur
                if (i.end < cur.start) continue;
                if (i.start > cur.end) continue;
                // find the max height beneath cur
                preMaxHeight = Math.max(preMaxHeight, i.height);
            }
            cur.height += preMaxHeight;
            intervals.add(cur);
            return cur.height;
        }
    https://blog.csdn.net/u014688145/article/details/78243313
    有点几何题的味道,实际上正方形往上叠总共就5种情况,如下: 
    alt text
    前四种情况,可以合并成一种情况处理,只需要搜索蓝色两条边界内是否存在对应的边,如果有,则说明需要叠加。第五种情况需要特殊处理,直接扫描所有正方形,包含蓝色矩形块则更新高度。
        public List<Integer> fallingSquares(int[][] positions) {
            int n = positions.length;
            List<Integer> ans = new ArrayList<>();
    
            TreeMap<Double, Integer> map = new TreeMap<>();
            List<Edge> heights = new ArrayList<Edge>();
            int max = 0;
    
            for (int i = 0; i < n; ++i) {
                int x = positions[i][0];
                int h = positions[i][1];
                int y = h + x - 1;
    
    
                int h_max = 0;
                for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) {
                    h_max = Math.max(h_max, map.get(e));
                }
    
                for (Edge edge : heights) {
                    if (edge.x <= x && edge.y >= y) {
                        h_max = Math.max(h_max, edge.h);
                    }
                }
    
                h_max += h;
    
                map.put(x * 1.0, h_max);
                map.put(y * 1.0, h_max);
                heights.add(new Edge(x, y, h_max));
    
                max = Math.max(max, h_max);
                ans.add(max);
            }
            return ans;
        }       


      public List<Integer> fallingSquares(int[][] positions) { int n = positions.length; List<Integer> ans = new ArrayList<>(); TreeMap<Double, Integer> map = new TreeMap<>(); for (int[] pos : positions) { int x = pos[0]; int h = pos[1]; int y = x + h - 1; map.put(x * 1.0, 0); map.put(y * 1.0, 0); } int max = 0; for (int i = 0; i < n; ++i) { int x = positions[i][0]; int h = positions[i][1]; int y = h + x - 1; int h_max = 0; for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) { h_max = Math.max(h_max, map.get(e)); } h_max += h; for (Double e : map.subMap(x - 0.1, y + 0.1).keySet()) { map.put(e, h_max); } max = Math.max(max, h_max); ans.add(max); } return ans; }
      X. Interval Tree
      https://leetcode.com/problems/falling-squares/discuss/108765/Java-14ms-beats-99.38-using-interval-tree
          class Node {
              public int l;
              public int r;
              public int max;
              public int height;
              public Node left;
              public Node right;
              
              public Node (int l, int r, int max, int height) {
                  this.l = l;
                  this.r = r;
                  this.max = max;
                  this.height = height;
              }
          }
          
          
          private boolean intersect(Node n, int l, int r) {
              if (r <= n.l || l >= n.r) {
                  return false;
              }
              return true;
          }
          
          private Node insert(Node root, int l, int r, int height) {
              if (root == null) {
                  return new Node(l, r, r, height);
              }
              
              if (l <= root.l) {
                  root.left = insert(root.left, l, r, height);
              } else {
                   // l > root.l
                  root.right = insert(root.right, l, r, height);
              }
              root.max = Math.max(r, root.max);
              return root;
          }
          
          // return the max height for interval (l, r)
          private int maxHeight(Node root, int l, int r) {
              if (root == null || l >= root.max) {
                  return 0;
              }
              
              int res = 0;
              if (intersect(root, l, r)) {
                  res = root.height;
              }
              if (r > root.l) {
                  res = Math.max(res, maxHeight(root.right, l, r));
              }
              res = Math.max(res, maxHeight(root.left, l, r));
              return res;
          }
          
          public List<Integer> fallingSquares(int[][] positions) {
              Node root = null;
              List<Integer> res = new ArrayList<>();
              int prev = 0;
              
              for (int i = 0; i < positions.length; ++i) {
                  int l = positions[i][0];
                  int r = positions[i][0] + positions[i][1];
                  int currentHeight = maxHeight(root, l, r);
                  root = insert(root, l, r, currentHeight + positions[i][1]);
                  prev = Math.max(prev, currentHeight + positions[i][1]);
                  res.add(prev);
              }
              
              return res;
          }



      No comments:

      Post a Comment

      Labels

      GeeksforGeeks (1109) LeetCode (1095) Review (846) Algorithm (795) to-do (634) LeetCode - Review (574) Classic Algorithm (324) Dynamic Programming (294) Classic Interview (288) Google Interview (242) Tree (145) POJ (139) Difficult Algorithm (132) LeetCode - Phone (127) EPI (125) Bit Algorithms (120) Different Solutions (120) Lintcode (113) Cracking Coding Interview (110) Smart Algorithm (109) Math (108) HackerRank (89) Binary Search (83) Binary Tree (82) Graph Algorithm (76) Greedy Algorithm (73) DFS (71) Stack (65) LeetCode - Extended (62) Interview Corner (61) List (58) Advanced Data Structure (56) BFS (54) Codility (54) ComProGuide (52) Algorithm Interview (50) Geometry Algorithm (49) Trie (49) Binary Search Tree (47) USACO (46) Interval (45) LeetCode Hard (42) Mathematical Algorithm (42) ACM-ICPC (41) Data Structure (40) Knapsack (40) Space Optimization (40) Jobdu (39) Matrix (39) Recursive Algorithm (39) String Algorithm (38) Union-Find (37) Backtracking (36) Codeforces (36) Introduction to Algorithms (36) Must Known (36) Beauty of Programming (35) Sort (35) Array (33) Data Structure Design (33) Segment Tree (33) Sliding Window (33) prismoskills (33) HDU (31) Priority Queue (31) Google Code Jam (30) Permutation (30) Puzzles (30) Array O(N) (29) Company-Airbnb (29) Company-Zenefits (28) Microsoft 100 - July (28) Palindrome (28) to-do-must (28) Graph (27) Random (27) Company - LinkedIn (25) GeeksQuiz (25) Logic Thinking (25) Post-Order Traverse (25) Pre-Sort (25) Time Complexity (25) hihocoder (25) Queue (24) Company-Facebook (23) High Frequency (23) TopCoder (23) Algorithm Game (22) Binary Indexed Trees (22) Bisection Method (22) Hash (22) DFS + Review (21) Lintcode - Review (21) Two Pointers (21) Brain Teaser (20) CareerCup (20) Company - Twitter (20) LeetCode - DP (20) Merge Sort (20) O(N) (20) Follow Up (19) UVA (19) Ordered Stack (18) Probabilities (18) Company-Uber (17) Game Theory (17) Topological Sort (17) Codercareer (16) Heap (16) Shortest Path (16) String Search (16) Tree Traversal (16) itint5 (16) Difficult (15) Iterator (15) BST (14) KMP (14) LeetCode - DFS (14) Number (14) Number Theory (14) Amazon Interview (13) Basic Algorithm (13) Codechef (13) Euclidean GCD (13) Long Increasing Sequence(LIS) (13) Majority (13) Reverse Thinking (13) mitbbs (13) Combination (12) Computational Geometry (12) LeetCode - Classic (12) Modify Tree (12) Reconstruct Tree (12) Reservoir Sampling (12) 尺取法 (12) AOJ (11) DFS+Backtracking (11) Fast Power Algorithm (11) Graph DFS (11) LCA (11) Miscs (11) Princeton (11) Proof (11) Rolling Hash (11) Tree DP (11) X Sum (11) 挑战程序设计竞赛 (11) Bisection (10) Bucket Sort (10) Coin Change (10) Company - Microsoft (10) DFS+Cache (10) Facebook Hacker Cup (10) HackerRank Easy (10) O(1) Space (10) SPOJ (10) Theory (10) TreeMap (10) Tutorialhorizon (10) DP-Multiple Relation (9) DP-Space Optimization (9) Divide and Conquer (9) Kadane - Extended (9) Mathblog (9) Max-Min Flow (9) Prefix Sum (9) Quick Sort (9) Simulation (9) Stack Overflow (9) Stock (9) System Design (9) Use XOR (9) Book Notes (8) Bottom-Up (8) Company-Amazon (8) DFS+BFS (8) Interval Tree (8) Left and Right Array (8) Linked List (8) Longest Common Subsequence(LCS) (8) Prime (8) Suffix Tree (8) Tech-Queries (8) Traversal Once (8) 穷竭搜索 (8) Algorithm Problem List (7) Expression (7) Facebook Interview (7) Fibonacci Numbers (7) Game Nim (7) Graph BFS (7) HackerRank Difficult (7) Hackerearth (7) Inversion (7) Kadane’s Algorithm (7) Level Order Traversal (7) Math-Divisible (7) Probability DP (7) Quick Select (7) Radix Sort (7) TreeSet (7) n00tc0d3r (7) 蓝桥杯 (7) Catalan Number (6) Classic Data Structure Impl (6) DFS+DP (6) DP - Tree (6) DP-Print Solution (6) Dijkstra (6) Dutch Flag (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Manacher (6) Minimum Spanning Tree (6) Morris Traversal (6) Multiple Data Structures (6) One Pass (6) Programming Pearls (6) Pruning (6) Rabin-Karp (6) Randomized Algorithms (6) Sampling (6) Schedule (6) Stream (6) Suffix Array (6) Threaded (6) Xpost (6) reddit (6) AI (5) Algorithm - Brain Teaser (5) Art Of Programming-July (5) Big Data (5) Brute Force (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Convex Hull (5) Crazyforcode (5) Cycle (5) DP-Include vs Exclude (5) Fast Slow Pointers (5) Find Rule (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Java (5) LeetCode - TODO (5) Matrix Chain Multiplication (5) Maze (5) Microsoft Interview (5) Pre-Sum (5) Quadtrees (5) Quick Partition (5) Quora (5) SPFA(Shortest Path Faster Algorithm) (5) Subarray Sum (5) Sudoku (5) Sweep Line (5) Word Search (5) jiuzhang (5) 单调栈 (5) 树形DP (5) 1point3acres (4) Abbreviation (4) Anagram (4) Anagrams (4) Approximate Algorithm (4) Backtracking-Include vs Exclude (4) Brute Force - Enumeration (4) Chess Game (4) Consistent Hash (4) Distributed (4) Eulerian Cycle (4) Flood fill (4) Graph-Classic (4) HackerRank AI (4) Histogram (4) Kadane Max Sum (4) Knapsack - Mixed (4) Knapsack - Unbounded (4) LeetCode - Recursive (4) MST (4) MinMax (4) N Queens (4) Nerd Paradise (4) Parallel Algorithm (4) Practical Algorithm (4) Probability (4) Programcreek (4) Spell Checker (4) Stock Maximize (4) Subset Sum (4) Subsets (4) Symbol Table (4) Triangle (4) Water Jug (4) algnotes (4) fgdsb (4) to-do-2 (4) 最大化最小值 (4) A Star (3) Algorithm - How To (3) Algorithm Design (3) B Tree (3) Big Data Algorithm (3) Caterpillar Method (3) Coins (3) Company - Groupon (3) Company - Indeed (3) Cumulative Sum (3) DP-Fill by Length (3) DP-Two Variables (3) Dedup (3) Dequeue (3) Dropbox (3) Easy (3) Finite Automata (3) Github (3) GoLang (3) Graph - Bipartite (3) Include vs Exclude (3) Joseph (3) Jump Game (3) K (3) Knapsack-多重背包 (3) LeetCode - Bit (3) Linked List Merge Sort (3) LogN (3) Master Theorem (3) Min Cost Flow (3) Minesweeper (3) Missing Numbers (3) NP Hard (3) O(N) Hard (3) Online Algorithm (3) Parser (3) Pascal's Triangle (3) Pattern Match (3) Project Euler (3) Rectangle (3) Scala (3) SegmentFault (3) Shuffle (3) Sieve of Eratosthenes (3) Skyline (3) Stack - Smart (3) State Machine (3) Subtree (3) Transform Tree (3) Trie + DFS (3) Two Pointers Window (3) Warshall Floyd (3) With Random Pointer (3) Word Ladder (3) bookkeeping (3) codebytes (3) Activity Selection Problem (2) Advanced Algorithm (2) AnAlgorithmADay (2) Application of Algorithm (2) Array Merge (2) BOJ (2) BT - Path Sum (2) Balanced Binary Search Tree (2) Bellman Ford (2) Binary Search - Smart (2) Binomial Coefficient (2) Bit Counting (2) Bit Mask (2) Bit-Difficult (2) Bloom Filter (2) Book Coding Interview (2) Branch and Bound Method (2) Clock (2) Codesays (2) Company - Baidu (2) Company-Snapchat (2) Complete Binary Tree (2) DFS+BFS, Flood Fill (2) DP - DFS (2) DP-3D Table (2) DP-Classical (2) DP-Output Solution (2) DP-Slide Window Gap (2) DP-i-k-j (2) DP-树形 (2) Distributed Algorithms (2) Divide and Conqure (2) Doubly Linked List (2) Edit Distance (2) Factor (2) Forward && Backward Scan (2) GoHired (2) Graham Scan (2) Graph BFS+DFS (2) Graph Coloring (2) Graph-Cut Vertices (2) Hamiltonian Cycle (2) Hard Algorithm (2) Huffman Tree (2) In-order Traverse (2) Include or Exclude Last Element (2) Information Retrieval (2) Interview - Linkedin (2) Invariant (2) Islands (2) Linked Interview (2) Linked List Sort (2) Longest SubArray (2) Lucene-Solr (2) Math-Remainder Queue (2) Matrix Power (2) Median (2) Minimum Vertex Cover (2) Negative All Values (2) Number Each Digit (2) Numerical Method (2) Object Design (2) Order Statistic Tree (2) Parent-Only Tree (2) Parentheses (2) Peak (2) Programming (2) Range Minimum Query (2) Regular Expression (2) Return Multiple Values (2) Reuse Forward Backward (2) Robot (2) Rosettacode (2) Scan from right (2) Search (2) SimHash (2) Simple Algorithm (2) Spatial Index (2) Strongly Connected Components (2) Summary (2) TV (2) Tile (2) Traversal From End (2) Tree Sum (2) Tree Traversal Return Multiple Values (2) Tree Without Tree Predefined (2) Word Break (2) Word Graph (2) Word Trie (2) Yahoo Interview (2) Young Tableau (2) 剑指Offer (2) 数位DP (2) 1-X (1) 51Nod (1) Akka (1) Algorithm - New (1) Algorithm Series (1) Algorithms Part I (1) Analysis of Algorithm (1) Array-Element Index Negative (1) Array-Rearrange (1) Augmented BST (1) Auxiliary Array (1) Auxiliary Array: Inc&Dec (1) BACK (1) BK-Tree (1) BZOJ (1) Basic (1) Bayes (1) Beauty of Math (1) Big Integer (1) Big Number (1) Binary (1) Binary Sarch Tree (1) Binary String (1) Binary Tree Variant (1) Bipartite (1) Bit-Missing Number (1) BitMap (1) BitMap index (1) BitSet (1) Bug Free Code (1) BuildIt (1) C/C++ (1) CC Interview (1) Cache (1) Calculate Height at Same Recusrion (1) Cartesian tree (1) Check Tree Property (1) Chinese (1) Circular Buffer (1) Clean Code (1) Cloest (1) Clone (1) Code Quality (1) Codesolutiony (1) Company - Alibaba (1) Company - Palantir (1) Company - WalmartLabs (1) Company-Apple (1) Company-Epic (1) Company-Salesforce (1) Company-Yelp (1) Compression Algorithm (1) Concise (1) Concurrency (1) Cont Improvement (1) Convert BST to DLL (1) Convert DLL to BST (1) Custom Sort (1) Cyclic Replacement (1) DFS-Matrix (1) DP - Probability (1) DP Fill Diagonal First (1) DP-Difficult (1) DP-End with 0 or 1 (1) DP-Fill Diagonal First (1) DP-Graph (1) DP-Left and Right Array (1) DP-MaxMin (1) DP-Memoization (1) DP-Node All Possibilities (1) DP-Optimization (1) DP-Preserve Previous Value (1) DP-Print All Solution (1) Database (1) Detect Negative Cycle (1) Diagonal (1) Directed Graph (1) Do Two Things at Same Recusrion (1) Domino (1) Dr Dobb's (1) Duplicate (1) Equal probability (1) External Sort (1) FST (1) Failure Function (1) Fraction (1) Front End Pointers (1) Funny (1) Fuzzy String Search (1) Game (1) Generating Function (1) Generation (1) Genetic algorithm (1) GeoHash (1) Geometry - Orientation (1) Google APAC (1) Graph But No Graph (1) Graph Transpose (1) Graph Traversal (1) Graph-Coloring (1) Graph-Longest Path (1) Gray Code (1) HOJ (1) Hanoi (1) How Hash (1) How to Test (1) Improve It (1) In Place (1) Inorder-Reverse Inorder Traverse Simultaneously (1) Interpolation search (1) Interview (1) Interview - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Knuth Shuffle (1) Kosaraju’s algorithm (1) Kruskal (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Construction (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode - Thinking (1) Linked List Reverse (1) Linkedin (1) Linkedin Interview (1) Local MinMax (1) Logic Pattern (1) Longest Common Subsequence (1) Longest Common Substring (1) Longest Prefix Suffix(LPS) (1) Machine Learning (1) Maintain State (1) Manhattan Distance (1) Map && Reverse Map (1) Math - Induction (1) Math-Multiply (1) Math-Sum Of Digits (1) Matrix - O(N+M) (1) Matrix BFS (1) Matrix Graph (1) Matrix Search (1) Matrix+DP (1) Matrix-Rotate (1) Max Min So Far (1) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-End BFS (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Element (1) Next Successor (1) Offline Algorithm (1) PAT (1) Parenthesis (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) PreProcess (1) Probabilistic Data Structure (1) Python (1) Queue & Stack (1) RSA (1) Ranking (1) Rddles (1) ReHash (1) Realtime (1) Recurrence Relation (1) Recursive DFS (1) Recursive to Iterative (1) Red-Black Tree (1) Region (1) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (1) Square (1) Streaming Algorithm (1) String Algorithm. Symbol Table (1) String DP (1) String Distance (1) SubMatrix (1) Subsequence (1) System of Difference Constraints(差分约束系统) (1) TSP (1) Ternary Search Tree (1) Test (1) Test Cases (1) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie and Heap (1) Trie vs Hash (1) Trie vs HashMap (1) Triplet (1) Two Data Structures (1) Two Stacks (1) Two-End-BFS (1) USACO - Classical (1) USACO - Problems (1) UyHiP (1) Valid Tree (1) Vector (1) Virtual Matrix (1) Wiggle Sort (1) Wikipedia (1) ZOJ (1) ZigZag (1) baozitraining (1) codevs (1) cos126 (1) javabeat (1) jum (1) namic Programming (1) sqrt(N) (1) 两次dijkstra (1) 九度 (1) 二进制枚举 (1) 夹逼法 (1) 归一化 (1) 折半枚举 (1) 枚举 (1) 状态压缩DP (1) 男人八题 (1) 英雄会 (1) 逆向思维 (1)

      Popular Posts