Saturday, October 15, 2016

LeetCode 414 - Third Maximum Number


https://leetcode.com/problems/third-maximum-number/
Given a non-empty array of integers, return the third maximum number in this array. If it does not exist, return the maximum number. The time complexity must be in O(n).
Example 1:
Input: [3, 2, 1]

Output: 1

Explanation: The third maximum is 1.
Example 2:
Input: [1, 2]

Output: 2

Explanation: The third maximum does not exist, so the maximum (2) is returned instead.
Example 3:
Input: [2, 2, 3, 1]

Output: 1

Explanation: Note that the third maximum here means the third maximum distinct number.
Both numbers with value 2 are both considered as second maximum.
https://discuss.leetcode.com/topic/63715/java-neat-and-easy-understand-solution-o-n-time-o-1-space
    public int thirdMax(int[] nums) {
        Integer max1 = null;
        Integer max2 = null;
        Integer max3 = null;
        for (Integer n : nums) {
            if (n.equals(max1) || n.equals(max2) || n.equals(max3)) continue;
            if (max1 == null || n > max1) {
                max3 = max2;
                max2 = max1;
                max1 = n;
            } else if (max2 == null || n > max2) {
                max3 = max2;
                max2 = n;
            } else if (max3 == null || n > max3) {
                max3 = n;
            }
        }
        return max3 == null ? max1 : max3;
    }
http://www.jianshu.com/p/e9dfd2494f41
 public int thirdMax (int[] nums) {
     long first = Long.MIN_VALUE;
     long second = Long.MIN_VALUE;
     long third = Long.MIN_VALUE;
     for (int i:nums){
         if (i>first){
             third = second;
             second = first;
             first = i;
         }else if (i == first)     //必要
             continue;
         else if (i > second){
             third = second;
             second = i;
         }else if (i == second)     //必要
             continue;
         else if (i > third){
             third = i;
         }
     }
     return third == Long.MIN_VALUE ? (int)first : (int)third;
       //method 要求返回类型为int,强制转化一下,narrowing conversion
 }
https://discuss.leetcode.com/topic/63671/o-n-time-o-1-space-java-short-solution
    public int thirdMax(int[] nums) {
        long first=Long.MIN_VALUE;
        long second=Long.MIN_VALUE;
        long third=Long.MIN_VALUE;
        for(int i:nums){
            if(i>first){
                third=second;
                second=first;
                first=i;
            }else if(i==first)
                continue;
            else if(i>second){
                third=second;
                second=i;
            }else if(i==second)
                continue;
            else if(i>third){
                third=i;
            }
        }
        return third==Long.MIN_VALUE?(int)first:(int)third;
    }

https://discuss.leetcode.com/topic/62236/java-solution-in-0ms-run-time-o-n-and-space-o-1
Another way to check whether the third maximum number exists is to declare all three numbers as long integer type and initialize them to the corresponding minimum value. Then at last if the third number is still the minimum value of long integer, we know it does not exist (its value will not get updated if it does not exist). Something like this:
public int thirdMax(int[] nums) {
    long max = Long.MIN_VALUE, mid = max, min = max;
        
    for (int ele : nums) {
        if (ele > max) {
            min = mid;
            mid = max;
            max = ele;
        } else if (max > ele && ele > mid) {
            min = mid;
            mid = ele;
        } else if (mid > ele && ele > min) {
            min = ele;
        }
    }
        
    return (int)(min != Long.MIN_VALUE ? min : max);
}

public int thirdMax(int[] nums) {
        int max, mid, small, count;
        max = mid = small = Integer.MIN_VALUE;
        count = 0;  //Count how many top elements have been found.

        for( int x: nums) {
            //Skip loop if max or mid elements are duplicate. The purpose is for avoiding right shift.
            if( x == max || x == mid ) {
                continue;
            }

            if (x > max) {
                //right shift
                small = mid;
                mid = max;

                max = x;
                count++;
            } else if( x > mid) {
                //right shift
                small = mid;

                mid = x;
                count++;
            } else if ( x >= small) { //if small duplicated, that's find, there's no shift and need to increase count.
                small = x;
                count++;
            }
        }

        //"count" is used for checking whether found top 3 maximum elements.
        if( count >= 3) { 
            return small;
        } else {
            return max;
        }
    }
http://bookshadow.com/weblog/2016/10/09/leetcode-third-maximum-number/
利用变量a, b, c分别记录数组第1,2,3大的数字
遍历一次数组即可,时间复杂度O(n)

def thirdMax(self, nums): """ :type nums: List[int] :rtype: int """ a = b = c = None for n in nums: if n > a: a, b, c = n, a, b elif a > n > b: b, c = n, b elif b > n > c: c = n return c if c is not None else a


X. TreeSet
https://discuss.leetcode.com/topic/63903/short-easy-c-using-set/8
  public int thirdMax(int[] nums) {
    TreeSet<Integer> set = new TreeSet<>();
    for(int num : nums) {
      set.add(num);
      if(set.size() > 3) {
        set.remove(set.first());
      }
    }
    return set.size() < 3 ? set.last() : set.first();
  }
https://discuss.leetcode.com/topic/65119/java-solution-using-treeset
    public final int N = 3;
    public int thirdMax(int[] nums) {
        if (nums.length == 0) return 0;

        TreeSet<Integer> set = new TreeSet<Integer>();
        for (int i = 0; i < nums.length; i++) {
            if (set.contains(nums[i])) continue;
            if (set.size() < N || nums[i] > set.first()) {
                if (set.size() == N) set.remove(set.first());
                set.add(nums[i]);
            }
        }
        return set.size() == N ? set.first() : set.last();
    }

http://www.cnblogs.com/grandyang/p/5983113.html
下面这种方法利用了set的自动排序和自动去重复项的特性,很好的解决了问题,对于遍历到的数字,加入set中,重复项就自动去掉了,如果此时set大小大于3个了,那么我们把set的第一个元素去掉,也就是将第四大的数字去掉,那么就可以看出set始终维护的是最大的三个不同的数字,最后遍历结束后,我们看set的大小是否为3,是的话就返回首元素,不是的话就返回尾元素

    int thirdMax(vector<int>& nums) {
        set<int> s;
        for (int num : nums) {
            s.insert(num);
            if (s.size() > 3) {
                s.erase(s.begin());
            }
        }
        return s.size() == 3 ? *s.begin() : *s.rbegin();
    }

X. Use PriorityQueue with HashSet to check duplicate
- Faster than TreeSet as there is no need to keep the elements in the PQ sorted as TreeSet
https://discuss.leetcode.com/topic/63086/java-priorityqueue-o-n-o-1


    public int thirdMax(int[] nums) {
        PriorityQueue<Integer> pq = new PriorityQueue<>();// min heap
        Set<Integer> set = new HashSet<>();
        for (int i : nums) {
            if (!set.contains(i)) {
                pq.offer(i);
                set.add(i);
                if (pq.size() > 3) {
                    set.remove(pq.poll());
                }
            }
        }
        if (pq.size() < 3) {// if there is less than 3, we return the the last one
            while (pq.size() > 1) {
                pq.poll();
            }
        }
        return pq.peek(); // otherwise kth is the first one in the min heap
    }
Related:
http://algorithms.tutorialhorizon.com/find-the-element-which-appears-maximum-number-of-times-in-the-array/
Objec­tive: Given an array of inte­gers, write a algo­rithm to find the ele­ment which appears max­i­mum num­ber of times in the array.
Bet­ter Solu­tion : Use Hashmap. Store the count of each ele­ment of array in a hash table and later check in Hash map which ele­ment has the max­i­mum count 

public void maxRepeatingElementUsingSorting(int [] arrA){
    if(arrA.length<1){
        System.out.println("Inavlid Array");
        return;
    }
    Arrays.sort(arrA);
    int count=1;
    int maxCount=1;
    int currentElement = arrA[0];
    int maxCountElement =arrA[0];
    for (int i = 1; i <arrA.length ; i++) {
        if(currentElement==arrA[i]){
            count++;
        }else{
            if(count>maxCount){
                maxCount = count;
                maxCountElement = currentElement;
            }
            currentElement = arrA[i];
            count = 1;
        }
    }
    System.out.println("Element repeating maximum no of times: " + maxCountElement + ", maximum count: " + maxCount);
}
Bet­ter Solu­tion (Con­di­tional) : O(n) time and O(1) extra space.
  • This solu­tion works only if array has pos­i­tive inte­gers and all the ele­ments in the array are in range from 0 to n-1 where n is the size of the array.
  • Nav­i­gate the array.
  • Update the array as for ith index :- arrA[arrA[i]% n] = arrA[arrA[i]% n] + n;
  • Now nav­i­gate the updated array and check which index has the max­i­mum value, that index num­ber is the ele­ment which has the max­i­mum occur­rence in the array.
public void MaxRepeatingElementInPlace(int [] arrA){
    int size = arrA.length;
    int maxCount=0;
    int maxIndex=0;
    for (int i = 0; i <size ; i++) {
        //get the index to be updated
        int index = arrA[i]% size;
        arrA[index] = arrA[index] + size;
    }
    for (int i = 0; i <size ; i++) {
        if(arrA[i]/size>maxCount){
            maxCount=arrA[i]/size;
            maxIndex=i;
        }
    }
    System.out.println("Element repeating maximum no of times: " + arrA[maxIndex]%size + ", maximum count: " + maxCount);
}

http://algorithms.tutorialhorizon.com/find-duplicates-in-an-given-array-in-on-time-and-o1-extra-space/
Given an array of inte­gers, find out dupli­cates in it.
Bet­ter Solu­tion : Use Hash map. Store the count of each ele­ment of array in a hash table and later check in Hash table if any ele­ment has count more than 1
Bet­ter Solu­tion (Con­di­tional) : O(n) time and O(1) extra space.
  • This solu­tion works only if array has pos­i­tive inte­gers and all the ele­ments in the array are in range from 0 to n-1 where n is the size of the array.
  • Nav­i­gate the array.
  • Update the array as for ith index :- arrA[arrA[i]] = arrA[arrA[i]]*-1 (if it already not negative).
  • If is already neg­a­tive, we have dupli­cates, return false.
Note:
  • The code given below does not han­dle the case when 0 is present in the array.
  • To han­dle 0 in array, while nav­i­gat­ing the array, when 0 is encoun­tered, replace it with INT_MIN and if INT_MIN is encoun­tered while tra­vers­ing, this means 0 is repeated in the array.
public void hasDuplicates(int[] arrA) {

    for (int i = 0; i < arrA.length; i++) {
        //check if element is negative, if yes the we have found the duplicate
        if (arrA[Math.abs(arrA[i])] < 0) {
            System.out.println("Array has duplicates : " + Math.abs(arrA[i]));
        } else {
            arrA[Math.abs(arrA[i])] = arrA[Math.abs(arrA[i])] * -1;
        }
    }
}


No comments:

Post a Comment

Labels

GeeksforGeeks (1107) LeetCode (993) Algorithm (795) Review (766) to-do (633) LeetCode - Review (514) Classic Algorithm (324) Dynamic Programming (293) Classic Interview (288) Google Interview (242) Tree (145) POJ (139) Difficult Algorithm (132) LeetCode - Phone (127) EPI (125) Different Solutions (120) Bit Algorithms (118) Lintcode (113) Cracking Coding Interview (110) Smart Algorithm (109) Math (107) HackerRank (89) Binary Tree (82) Binary Search (81) Graph Algorithm (74) Greedy Algorithm (72) DFS (67) LeetCode - Extended (62) Interview Corner (61) Stack (60) List (58) Advanced Data Structure (56) BFS (54) Codility (54) ComProGuide (52) Algorithm Interview (50) Geometry Algorithm (48) Binary Search Tree (46) USACO (46) Trie (45) Mathematical Algorithm (42) ACM-ICPC (41) Interval (41) Data Structure (40) Knapsack (40) Space Optimization (40) Jobdu (39) LeetCode Hard (39) Recursive Algorithm (39) Matrix (38) String Algorithm (38) Backtracking (36) Codeforces (36) Introduction to Algorithms (36) Must Known (36) Beauty of Programming (35) Sort (35) Union-Find (34) Array (33) prismoskills (33) Segment Tree (32) Sliding Window (32) Data Structure Design (31) HDU (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) Priority Queue (27) Random (27) Graph (26) Company - LinkedIn (25) GeeksQuiz (25) Logic Thinking (25) Pre-Sort (25) hihocoder (25) Queue (24) Company-Facebook (23) High Frequency (23) Post-Order Traverse (23) TopCoder (23) Algorithm Game (22) Bisection Method (22) Hash (22) Binary Indexed Trees (21) DFS + Review (21) Lintcode - Review (21) Brain Teaser (20) CareerCup (20) Company - Twitter (20) Merge Sort (20) O(N) (20) Follow Up (19) Time Complexity (19) Two Pointers (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) Number (14) Number Theory (14) Amazon Interview (13) Basic Algorithm (13) Codechef (13) Euclidean GCD (13) KMP (13) Long Increasing Sequence(LIS) (13) Majority (13) mitbbs (13) Combination (12) Computational Geometry (12) LeetCode - Classic (12) Modify Tree (12) Reconstruct Tree (12) Reservoir Sampling (12) Reverse Thinking (12) 尺取法 (12) AOJ (11) DFS+Backtracking (11) Fast Power Algorithm (11) Graph DFS (11) LCA (11) LeetCode - DFS (11) Miscs (11) Princeton (11) Proof (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) Rolling Hash (10) SPOJ (10) Theory (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) TreeMap (9) Use XOR (9) Book Notes (8) Bottom-Up (8) Company-Amazon (8) DFS+BFS (8) LeetCode - DP (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) Interval Tree (7) Inversion (7) Kadane’s Algorithm (7) Level Order Traversal (7) Math-Divisible (7) Probability DP (7) Quick Select (7) Radix Sort (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) TreeSet (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) 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) LeetCode - TODO (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) Pascal's Triangle (3) Pattern Match (3) Project Euler (3) Rectangle (3) Scala (3) SegmentFault (3) Shuffle (3) Sieve of Eratosthenes (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) 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) Parser (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) Skyline (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) 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) 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) Hard Algorithm (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 - Detail (1) LeetCode - Related (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) 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