Friday, December 18, 2015

Get all palidromes after delete - Linkedin


给一个string, 可以删除任意字符,求所有可以得到的palidrome字串集
http://www.mitbbs.com/article_t/JobHunting/33053715.html
private static ISet<string> getPal(string s){
  IDictionary<char, IList<int>> dict = new Dictionary<char, IList<int>>();
  for(int i = 0; i < s.Length; i++){
    if(!dict.ContainsKey(s[i])) dict[s[i]] = new List<int>();
    dict[s[i]].Add(i);
  }
  ISet<string> result = new HashSet<string>();
  dfs(result, new StringBuilder(), new StringBuilder(), s, 0, s.Length - 1, 
dict);
  return result;
}
private static void dfs(ISet<string> result, StringBuilder sLeft, 
StringBuilder sRight, string s, int left, int right, IDictionary<char, IList
<int>> dict){
  for(int i = left; i <= right; i++){
    result.Add(sLeft.ToString() + s[i] + sRight.ToString());
    for(int j = 0; j < dict[s[i]].Count; j++)
      if(dict[s[i]][j] <= i) continue;
      else if (dict[s[i]][j] > right) break;
      else{
        sLeft.Append(s[i]);
        sRight.Insert(0, s[i]);
        result.Add(sLeft.ToString() + sRight.ToString());
        dfs(result, sLeft, sRight, s, i + 1, dict[s[i]][j] - 1, dict);
        sLeft.Length -= 1;
        sRight.Remove(0, 1);
        dfs(result, sLeft, sRight, s, i + 1, dict[s[i]][j] - 1, dict);
      }
  }
}

function reverse(s) {
    var o = '';
    for (var i = s.length - 1; i >= 0; i--)
        o += s[i];
    return o;
};

var p = function(s) {
    var prevLookup = {};
    var nextLookup = {};
    
    for (var i = 0; i < s.length; i++) {
        var c = s.charAt(i);
        if (!prevLookup[c]) {
            prevLookup[c] = [];
            nextLookup[c] = [];
        }
    }

    for (var i = 0; i < s.length; i++) {
        for(var c in prevLookup) {
            var lookup = prevLookup[c];
            if (c == s.charAt(i)) {
                lookup[i] = i;
            } else {
                if (i == 0) {
                    lookup[i] = null;
                } else {
                    lookup[i] = lookup[i-1];
                }
            }
        }
    }

    for (var i = s.length-1; i >= 0; i--) {
        for(var c in nextLookup) {
            var lookup = nextLookup[c];
            if (c == s.charAt(i)) {
                lookup[i] = i;
            } else {
                if (i == s.length-1) {
                    lookup[i] = null;
                } else {
                    lookup[i] = lookup[i+1];
                }
            }
        }
    }

    console.log(prevLookup);
    console.log(nextLookup);

    var _rec = function(left, right, accum) {
        console.log(accum + reverse(accum));
        if (left > right) return;


        for (var c in prevLookup) {
            var leftMostPos = nextLookup[c][left];
            var rightMostPos = prevLookup[c][right];
            if (leftMostPos != null && leftMostPos <= right) {
                console.log(accum + c + reverse(accum));
                if (rightMostPos > leftMostPos) {
                    _rec(leftMostPos + 1, rightMostPos -1, accum + c);
                }
            }
        }
    };

    _rec(0, s.length-1, "");
};

var s = "abcbabcba";
var s2 = "aaaaaaaa";
无重复的逐一完全遍历所有的substring 
palindrome。复杂度就是O(the number of palindrome substrings).
blaze 的解法很巧妙。关键是建立两个不同方向的lookup table,每个entry分别是字符
以及一个数组,数组中的 i 元素 是从 i 位置往左(右)看能看到的该字符的最近位
置。然后从两边向中间扫描。

关于leetcode 266, 267 参考:

http://buttercola.blogspot.com/2015/08/leetcode-palindrome-perm
very nice solution
关键是对每个可能的char,维护了一维数组,数组中的i意思是从这个i看过去下一个
same char出现的位置,这样DFS就是对char来做遍历。
而且这个解法很巧妙的解决了奇数长和偶数长的问题。
我的java实现,参考了Blaze的思路,不过不知道真正理解了他的思路没有,因为我不
太看得懂javascript.下面是几个点:

1. 这个题目用的是backtracking而不是dp,因为dp不可能给你产生排列组合。dp通常
就是给你一个最值。
2. 时间复杂度是指数级。
3. 题目的本质是产生排列。
4. 一般的backtracking来形成字符串的话,从现有字母集中每次任意挑一个字母,然后
从剩下的字母里继续挑。把所有字母都加到字符串里面去了,这轮程序也就完了。这个
题目因为有个顺序的限制条件,决定程序怎样进行下去并不能用剩下多少字符来决定,
而应该是用已经加入字符串的字母的位置来决定。


import java.util.HashMap;
import java.util.List;
import java.util.ArrayList;
import java.util.Iterator;
import java.util.Set;

public class Example5 {
    public static void main(String [] args){
        Example5 o = new Example5();
        List <String> results = o.setUp("abcbabcba");
        for(String str: results){
            System.out.println(str);
        }
    }
    public List <String> setUp(String palidrome){
        HashMap <Character, List<Integer>> map = new HashMap <Character, 
List<Integer>>();
        HashMap <Character, Pointers> locations = new HashMap <Character, 
Pointers>();
        char [] chs = palidrome.toCharArray();
        int i = 0;
        for(i=0;i<chs.length;i++){
            if(map.containsKey(chs[i])){
                map.get(chs[i]).add(i);
            }
            else{
                List <Integer> tmp = new ArrayList <Integer> ();
                tmp.add(i);
                map.put(chs[i],tmp);
            }
        }
        Set <Character> keySet = map.keySet();
        Iterator <Character> iter = keySet.iterator();
        List <String > results = new ArrayList <String> ();
        while(iter.hasNext()){
            Character ch=iter.next();
            //results.add(ch.toString());
            int size = map.get(ch).size();
            locations.put(ch, new Pointers(0,size-1));
        }
        StringBuilder sb1 = new StringBuilder();
        StringBuilder sb2 = new StringBuilder();
        dfs(results,0,chs.length-1,sb1,sb2,keySet,map,locations);
        return results;                    
    }
    
    private class Pointers {
        int left;
        int right;
        public Pointers(int left, int right){
            this.left = left;
            this.right = right;
        }
    }
    
    public void dfs (List <String> results, int left, int right, 
StringBuilder sb1, StringBuilder sb2, Set <Character> keySet, HashMap <
Character, List<Integer>> map, HashMap <Character, Pointers> locations) {
        Iterator <Character> iter = keySet.iterator();
        while(iter.hasNext()){
            Character ch=iter.next();
            //System.out.println("ch="+ch+" left="+left+" right="+right);
            Pointers p=locations.get(ch);
            //System.out.println("left1="+p.left+" right1="+p.right);
            List <Integer> nums = map.get(ch);
            int oldLeft = p.left;
            int oldRight = p.right;
            while((p.left<=p.right)&&(nums.get(p.left)<left)){
                p.left++;
            }
            while((p.left<=p.right)&&(nums.get(p.right)>right)){
                p.right--;
            }
            if(p.left<=p.right){
                results.add(sb1.toString()+ch+sb2.toString());
                //System.out.println(sb1.toString()+ch+sb2.toString());
            }
            if(p.left<p.right){
                sb1.append(ch);
                sb2.insert(0,ch);
                int tmp1 = p.left;
                int tmp2 = p.right;
                p.left++;
                p.right--;
                //System.out.println(sb1.toString()+sb2.toString());
                results.add(sb1.toString()+sb2.toString());
                dfs(results,nums.get(tmp1)+1,nums.get(tmp2)-1,sb1,sb2,keySet
,map,locations);
                sb1.deleteCharAt(sb1.length()-1);
                sb2.deleteCharAt(0);                    
            }
            p.left = oldLeft;
            p.right = oldRight;
        }
    }
}
这所有的回文字串有点难啊,我把mitbbs上的解法翻译成C++了. 1point3acres.com/bbs

#include <vector>
#include <iostream>
#include <string>
#include <map>. From 1point 3acres bbs
using namespace std;
鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
map<char, vector<int>> leftbook;
map<char, vector<int>> rightbook;

void fill(string s) {
    for (auto x : s) {
        if (leftbook.find(x) == leftbook.end())
            leftbook[x] = vector<int>(s.size(), -1);
        if (rightbook.find(x) == rightbook.end())
            rightbook[x] = vector<int>(s.size(), -1);
    }. visit 1point3acres.com for more.

    for (int i = 0; i < s.size(); ++i) {
        for (auto p = leftbook.begin(); p != leftbook.end(); ++p) {
            if (s == p->first)
                (p->second) = i;
            else 鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
                (p->second) = (i == 0 ? -1 : (p->second)[i-1]);
        }.鐣欏璁哄潧-涓€浜�-涓夊垎鍦�
    }. From 1point 3acres bbs

    for (int i = s.size() - 1; i >= 0; --i) {
        for (auto p = rightbook.begin(); p != rightbook.end(); ++p) {
            if (s == p->first). more info on 1point3acres.com
                (p->second) = i;
            else
                (p->second) = (i == s.size() - 1 ? -1 : (p->second)[i+1]);
        }
    }

}
-google 1point3acres
void fun(int l, int r, string half) {
    cout<<half + string(half.rbegin(), half.rend())<<endl;
    for (auto p = leftbook.begin(); p != leftbook.end(); ++p) {.鐣欏璁哄潧-涓€浜�-涓夊垎鍦�
        char c = p->first;
        int left = rightbook[c][l];
        int right = leftbook[c][r];
        if (left != -1 && left <= r) {.鏈枃鍘熷垱鑷�1point3acres璁哄潧
            cout<<half + c + string(half.rbegin(), half.rend())<<endl;
            if (right > left)
                fun(left + 1, right - 1, half + c);
        }
    }.鐣欏璁哄潧-涓€浜�-涓夊垎鍦�
}. visit 1point3acres.com for more.
鏉ユ簮涓€浜�.涓夊垎鍦拌鍧�. 
int main(int argc, char const *argv[])
{
    string s = "abcba";
    fill(s);
    fun(0, s.size() - 1, "");
    return 0;. from: 1point3acres.com/bbs 
}
补充内容 (2015-11-6 11:29):
leftbook是每个字符往左能看到的位置,rightbook是每个字符往右能看到的位置。
算法就是对所有的字符遍历并拼接起来。拼接的条件是,下一个字符必须落在上一个字符的两个book所规定的范围里



No comments:

Post a Comment

Labels

GeeksforGeeks (959) Algorithm (811) LeetCode (639) to-do (598) Review (343) Classic Algorithm (334) Classic Interview (299) Dynamic Programming (263) Google Interview (233) LeetCode - Review (229) Tree (146) POJ (137) Difficult Algorithm (136) EPI (127) Different Solutions (118) Bit Algorithms (110) Cracking Coding Interview (110) Smart Algorithm (109) Math (91) HackerRank (85) Lintcode (83) Binary Search (73) Graph Algorithm (73) Greedy Algorithm (61) Interview Corner (61) List (58) Binary Tree (56) DFS (56) Algorithm Interview (53) Advanced Data Structure (52) Codility (52) ComProGuide (52) LeetCode - Extended (47) USACO (46) Geometry Algorithm (45) BFS (43) Data Structure (42) Mathematical Algorithm (42) ACM-ICPC (41) Interval (38) Jobdu (38) Recursive Algorithm (38) Stack (38) String Algorithm (38) Binary Search Tree (37) Knapsack (37) Codeforces (36) Introduction to Algorithms (36) Matrix (36) Must Known (36) Beauty of Programming (35) Sort (35) Array (33) Trie (33) prismoskills (33) Segment Tree (32) Space Optimization (32) Union-Find (32) Backtracking (31) HDU (31) Google Code Jam (30) Permutation (30) Puzzles (30) Array O(N) (29) Data Structure Design (29) Company-Zenefits (28) Microsoft 100 - July (28) to-do-must (28) Random (27) Sliding Window (26) GeeksQuiz (25) Logic Thinking (25) hihocoder (25) High Frequency (23) Palindrome (23) Algorithm Game (22) Company - LinkedIn (22) Graph (22) Queue (22) DFS + Review (21) Hash (21) TopCoder (21) Binary Indexed Trees (20) Brain Teaser (20) CareerCup (20) Company - Twitter (20) Pre-Sort (20) Company-Facebook (19) UVA (19) Probabilities (18) Follow Up (17) Codercareer (16) Company-Uber (16) Game Theory (16) Heap (16) Shortest Path (16) String Search (16) Topological Sort (16) Tree Traversal (16) itint5 (16) Iterator (15) Merge Sort (15) O(N) (15) Bisection Method (14) Difficult (14) Number (14) Number Theory (14) Post-Order Traverse (14) Priority Quieue (14) Amazon Interview (13) BST (13) Basic Algorithm (13) Codechef (13) Majority (13) mitbbs (13) Combination (12) Computational Geometry (12) KMP (12) Long Increasing Sequence(LIS) (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) LeetCode - DFS (11) Ordered Stack (11) Princeton (11) Tree DP (11) 挑战程序设计竞赛 (11) Binary Search - Bisection (10) Company - Microsoft (10) Company-Airbnb (10) Euclidean GCD (10) Facebook Hacker Cup (10) HackerRank Easy (10) Reverse Thinking (10) Rolling Hash (10) SPOJ (10) Theory (10) Tutorialhorizon (10) X Sum (10) Coin Change (9) Lintcode - Review (9) Mathblog (9) Max-Min Flow (9) Stack Overflow (9) Stock (9) Two Pointers (9) Book Notes (8) Bottom-Up (8) DP-Space Optimization (8) Divide and Conquer (8) Graph BFS (8) LeetCode - DP (8) LeetCode Hard (8) Prefix Sum (8) Prime (8) System Design (8) Tech-Queries (8) Time Complexity (8) Use XOR (8) 穷竭搜索 (8) Algorithm Problem List (7) DFS+BFS (7) Facebook Interview (7) Fibonacci Numbers (7) Game Nim (7) HackerRank Difficult (7) Hackerearth (7) Interval Tree (7) Linked List (7) Longest Common Subsequence(LCS) (7) Math-Divisible (7) Miscs (7) O(1) Space (7) Probability DP (7) Radix Sort (7) Simulation (7) Suffix Tree (7) Xpost (7) n00tc0d3r (7) 蓝桥杯 (7) Bucket Sort (6) Catalan Number (6) Classic Data Structure Impl (6) DFS+DP (6) DP - Tree (6) How To (6) Interviewstreet (6) Knapsack - MultiplePack (6) Level Order Traversal (6) Manacher (6) Minimum Spanning Tree (6) One Pass (6) Programming Pearls (6) Quick Select (6) Rabin-Karp (6) Randomized Algorithms (6) Sampling (6) Schedule (6) Suffix Array (6) Threaded (6) reddit (6) AI (5) Art Of Programming-July (5) Big Data (5) Brute Force (5) Code Kata (5) Codility-lessons (5) Coding (5) Company - WMware (5) Crazyforcode (5) DFS+Cache (5) DP-Multiple Relation (5) DP-Print Solution (5) Dutch Flag (5) Fast Slow Pointers (5) Graph Cycle (5) Hash Strategy (5) Immutability (5) Inversion (5) Java (5) Kadane - Extended (5) Kadane’s Algorithm (5) Matrix Chain Multiplication (5) Microsoft Interview (5) Morris Traversal (5) Pruning (5) Quadtrees (5) Quick Partition (5) Quora (5) SPFA(Shortest Path Faster Algorithm) (5) Subarray Sum (5) Sweep Line (5) Traversal Once (5) TreeMap (5) jiuzhang (5) to-do-2 (5) 单调栈 (5) 树形DP (5) 1point3acres (4) Anagram (4) Approximate Algorithm (4) Backtracking-Include vs Exclude (4) Brute Force - Enumeration (4) Chess Game (4) Company-Amazon (4) Consistent Hash (4) Convex Hull (4) Cycle (4) DP-Include vs Exclude (4) Dijkstra (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) Left and Right Array (4) MinMax (4) Multiple Data Structures (4) N Queens (4) Nerd Paradise (4) Parallel Algorithm (4) Practical Algorithm (4) Pre-Sum (4) Probability (4) Programcreek (4) Quick Sort (4) Spell Checker (4) Stock Maximize (4) Subsets (4) Sudoku (4) Symbol Table (4) TreeSet (4) Triangle (4) Water Jug (4) Word Ladder (4) algnotes (4) fgdsb (4) 最大化最小值 (4) A Star (3) Abbreviation (3) Algorithm - Brain Teaser (3) Algorithm Design (3) Anagrams (3) B Tree (3) Big Data Algorithm (3) Binary Search - Smart (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) Edit Distance (3) Expression (3) Finite Automata (3) Forward && Backward Scan (3) Github (3) GoLang (3) Include vs Exclude (3) Joseph (3) Jump Game (3) Knapsack-多重背包 (3) LeetCode - Bit (3) LeetCode - TODO (3) Linked List Merge Sort (3) LogN (3) Master Theorem (3) Maze (3) Min Cost Flow (3) Minesweeper (3) Missing Numbers (3) NP Hard (3) Online Algorithm (3) Pascal's Triangle (3) Pattern Match (3) Project Euler (3) Rectangle (3) Scala (3) SegmentFault (3) Stack - Smart (3) State Machine (3) Streaming Algorithm (3) Subset Sum (3) Subtree (3) Transform Tree (3) Two Pointers Window (3) Warshall Floyd (3) With Random Pointer (3) Word Search (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) Binomial Coefficient (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) 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) GoHired (2) Graham Scan (2) Graph - Bipartite (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) Knuth Shuffle (2) LeetCode - Recursive (2) Linked Interview (2) Linked List Sort (2) Longest SubArray (2) Lucene-Solr (2) MST (2) MST-Kruskal (2) Math-Remainder Queue (2) Matrix Power (2) Minimum Vertex Cover (2) Negative All Values (2) Number Each Digit (2) Numerical Method (2) Object Design (2) Order Statistic Tree (2) Palindromic (2) Parentheses (2) Parser (2) Peak (2) Programming (2) Range Minimum Query (2) Reuse Forward Backward (2) Robot (2) Rosettacode (2) Scan from right (2) Search (2) Shuffle (2) Sieve of Eratosthenes (2) SimHash (2) Simple Algorithm (2) Skyline (2) Spatial Index (2) Stream (2) Strongly Connected Components (2) Summary (2) TV (2) Tile (2) Traversal From End (2) Tree Sum (2) Tree Traversal Return Multiple Values (2) Word Break (2) Word Graph (2) Word Trie (2) Young Tableau (2) 剑指Offer (2) 数位DP (2) 1-X (1) 51Nod (1) Akka (1) Algorithm - How To (1) Algorithm - New (1) Algorithm Series (1) Algorithms Part I (1) Analysis of Algorithm (1) Array-Element Index Negative (1) Array-Rearrange (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 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) Code Quality (1) Codesolutiony (1) Company - Alibaba (1) Company - Palantir (1) Company - WalmartLabs (1) Company-Apple (1) Company-Epic (1) Company-Salesforce (1) Company-Snapchat (1) Company-Yelp (1) Compression Algorithm (1) Concurrency (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) 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 - Easy (1) Interview - Facebook (1) Isomorphic (1) JDK8 (1) K Dimensional Tree (1) Knapsack - Fractional (1) Knapsack - ZeroOnePack (1) Knight (1) Kosaraju’s algorithm (1) Kruskal (1) Kruskal MST (1) Kth Element (1) Least Common Ancestor (1) LeetCode - Binary Tree (1) LeetCode - Coding (1) LeetCode - Detail (1) LeetCode - Related (1) LeetCode Diffcult (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) 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) Median (1) Memory-Efficient (1) MinHash (1) MinMax Heap (1) Monotone Queue (1) Monto Carlo (1) Multi-Reverse (1) Multiple DFS (1) Multiple Tasks (1) Next Successor (1) Offline Algorithm (1) PAT (1) Partition (1) Path Finding (1) Patience Sort (1) Persistent (1) Pigeon Hole Principle (1) Power Set (1) Pratical Algorithm (1) Probabilistic Data Structure (1) Proof (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) Regular Expression (1) Resources (1) Reverse Inorder Traversal (1) Robin (1) Selection (1) Self Balancing BST (1) Similarity (1) Sort && Binary Search (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) Thread (1) TimSort (1) Top-Down (1) Tournament (1) Tournament Tree (1) Transform Tree in Place (1) Tree Diameter (1) Tree Rotate (1) Trie + DFS (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) Wiggle Sort (1) Wikipedia (1) Yahoo Interview (1) ZOJ (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