06 2015 档案
摘要：Each time we find a match, increase the global counter by 1.For KMP, algorithm, you may refer to the following links which have nice explanations.KMP ...
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摘要：Well, this problem becomes a little trickier since there may be more than one majority element. But, there can be at most two of them. This can be pro...
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摘要：Well, if you have got this problem accepted, you may have noticed that there are 7 suggested solutions for this problem. The following passage will im...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param dictionary: a vector of strings 5 * @return: a vector of strings 6 */ 7 vector l...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param string: An array of Char 5 * @param length: The true length of the string 6 * @retur...
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摘要：暴力解法（O(mn)）： 1 class Solution { 2 public: 3 /** 4 * Returns a index to the first occurrence of target in source, 5 * or -1 if target is...
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摘要：1 /** 2 * Definition of ListNode 3 * class ListNode { 4 * public: 5 * int val; 6 * ListNode *next; 7 * ListNode(int val) { 8 * ...
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摘要：Well, life gets difficult pretty soon whenever the same operation on array is transferred to linked list.First, a quick recap of insertion sort:Start ...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param A: sorted integer array A which has m elements, 5 * but size of A is m+n 6 ...
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摘要：The recursive solution is trivial and I omit it here.Iterative Solution using Stack (O(n)time andO(n)space): 1 /** 2 * Definition of TreeNode: 3 * c...
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摘要：A subroutine of merge sort. 1 class Solution { 2 public: 3 /** 4 * @param A and B: sorted integer array A and B. 5 * @return: A new sort...
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摘要：The recursive solution is trivial and I omit it here.Iterative Solution using Stack (O(n)time andO(n)space): 1 /** 2 * Definition of TreeNode: 3 * c...
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摘要：The recursive solution is trivial and I omit it here.Iterative Solution using Stack (O(n) time and O(n) space): 1 /** 2 * Definition of TreeNode: 3 ...
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摘要：动态规划：lis[i] = max_{j = 0, 1, ..., i - 1, nums[j] nums) { 8 // write your code here 9 vector lis(nums.size(), 1);10 int maxlen...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param strs: A list of strings 5 * @return: The longest common prefix 6 */ 7 string...
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摘要：Bit-by-Bit summation: 1 class Solution { 2 public: 3 /* 4 * @param a: The first integer 5 * @param b: The second integer 6 * @retur...
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摘要：The hints of the problem has given detailed information to implement the topological sorting algorithm. In fact, the toposort algorithm of the hint is...
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摘要：BFS: 1 /** 2 * Definition for Directed graph. 3 * struct DirectedGraphNode { 4 * int label; 5 * vector neighbors; 6 * DirectedGraphNo...
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摘要：基于快速排序： 1 class Solution { 2 public: 3 /* 4 * param k : description of k 5 * param nums : description of array and index 0 ~ n-1 6 ...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param grid: a list of lists of integers. 5 * @return: An integer, minimizes the sum of all numb...
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摘要：Well, to compute the number of trailing zeros, we need to first think clear about what will generate a trailing0? Obviously, a number multiplied by10w...
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摘要：1 class Solution { 2 public: 3 // param n : description of n 4 // return: description of return 5 long long trailingZeros(long long n) ...
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摘要：Well, the basic idea is very simple. Start from the tail ofsand move backwards to find the first non-space character. Then from this character, move b...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param s A string 5 * @return the length of last word 6 */ 7 int lengthOfLastWord(strin...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param s A string 5 * @return Whether the string is a valid palindrome 6 */ 7 bool isPa...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param s A string 5 * @return whether the string is a valid parentheses 6 */ 7 bool isV...
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摘要：递归实现： 1 /** 2 * Definition of TreeNode: 3 * class TreeNode { 4 * public: 5 * int val; 6 * TreeNode *left, *right; 7 * TreeNode(int v...
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摘要：1 class MinStack { 2 public: 3 MinStack() { 4 // do initialization if necessary 5 } 6 7 void push(int number) { 8 // wri...
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摘要：1 class Queue { 2 public: 3 stack stack1; 4 stack stack2; 5 6 Queue() { 7 // do intialization if necessary 8 } 9 10 void...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param A, B: Two strings. 5 * @return: The length of longest common subsequence of A and B. 6 ...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param A, B: Two string. 5 * @return: the length of the longest common substring. 6 */ ...
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摘要：Well, this problem is spiritually similar to toCourse Schedule. You only need to store the nodes in the order you visit into a vector during BFS or DF...
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摘要：As suggested by the hints, this problem is equivalent to detecting a cycle in the graph represented byprerequisites. Both BFS and DFS can be used to s...
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摘要：Topological sort is an important application of DFS in directed acyclic graphs (DAG). For each edge (u, v) from node u to node v in the graph, u must ...
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摘要：Graph is an important data structure and has many important applications. Moreover, grach traversal is key to many graph algorithms.There are two syst...
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摘要：Problem Description:Given two strings S and T, determine if they are both one edit distance apart.To solve this problem, you first need to know what i...
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摘要：This problem is similar to Missing Ranges and easier than that one.The idea is to use two pointers to find the beginning and end of a range and then p...
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摘要：Problem Description:Given a sorted integer array where the range of elements are [lower,upper] inclusive, return its missing ranges.For example, given...
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摘要：Well, the key to this problem is on how to identify the recurring parts. After doing some examples using pen and paper, you may find that for the deci...
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摘要：The problem statement has stated that there are bothO(n)andO(nlogn)solutions to this problem. Let's see theO(n)solution first (taken fromthis link), w...
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摘要：Problem Description:Given an input string, reverse the string word by word. A word is defined as a sequence of non-space characters.The input string d...
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摘要：This link has a great discussion about this problem. You may refer to it if you like. In fact, the idea and code in this passage is from the former li...
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摘要：Well, this problem has a nice BFS structure.Let's see the example in the problem statement.start = "hit"end = "cog"dict = ["hot", "dot", "dog", "lot",...
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摘要：This problem is not quite difficult (a typical BFS traversal of graph), though, its aceptance rate is relatively low. In fact, the key obstacle in pas...
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摘要：This problem is an application of graph traversal, which has two systematic methods:Bread-First Search (BFS)andDepth-First Search (DFS). In the follow...
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摘要：To solve this problem, some observations have to be made first.Let's first see two relatively easy observations.To maximize the probability that we ca...
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摘要：This is a typical problem aboutsearching. In fact, you can use either BFS or DFS for it. Personally, I use BFS because I think it is more intuitive an...
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摘要：The basic idea of is as follows:Maintain a dequeoperandsfor the numbers and another dequeoperationsfor the operators+, -, *,/`.Scan the expression fro...
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摘要：This problem can be solved elegantly using dynamic programming.We maintain two arrays:cnt[i][j] --- number of parentheses needed to add within s[i..j]...
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摘要：KMP is a classic and yet notoriously hard-to-understand algorithm. However, I think the following two links give nice explanations. You may refer to t...
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摘要：Well, a typical backtracking problem. Make sure you are clear with the following three problems:What is a partial solution and when is it finished? --...
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摘要：Well, a typical backtracking problem. Make sure you are clear with the following three problems:What is a partial solution and when is it finished? --...
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摘要：A typical backtracking problem. For any backtracking problem, you need to be think about three ascepts:What is a partial solution and when is it finis...
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摘要：The Longest Increasing Subsequence (LIS) problem requires us to find a subsequence t of a given sequence s, such that t satisfies two requirements:Ele...
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摘要：This is the classic LCS problem. Since it only requires you to print the maximum length, the code can be optimized to use only O(m) space (seehere).My...
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摘要：This is the classic LCS problem. Since it requires you to print one longest common subsequence, just use the O(m*n)-space version here.My accepted cod...
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摘要：The Longest Common Subsequence (LCS) problem is as follows:Given two sequences s andt, find the length of the longest sequence r, which is a subsequen...
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摘要：The Longest Common Substring (LCS) problem is as follows:Given two strings s and t, find the length of the longest string r, which is a substring of b...
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摘要：This problem is a nice extension of the Valid Parentheses problem.There are several ways to solve it. The first idea is also to use a stack. However, ...
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摘要：Well, there are two ways to add a open or close parenthesis to the current string.If number of(is less thann, you can add(;If number of)is less than n...
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摘要：This is a classic problem of the application of stacks. The idea is, each time we meet a(,{or[, we push it to a stack. If we meet a),}or], we check if...
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摘要：Problem Description:Given a binary tree where all the right nodes are either leaf nodes with a sibling (a left node that shares the same parent node) ...
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摘要：Problem Description:Given a string, find the length of the longest substring T that contains at most 2 distinct characters.For example, Given s =“eceb...
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摘要：After reading the quote below the problem, you will be motivated to solve it immediately :-) Well, indeed it is also relative easy. The process of inv...
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摘要：This problem is not very intuitive at first glance. However, the final idea should be very self-explanatory. You visit each element in nums, and then ...
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摘要：The idea is to find the longest palindromic substring ofsthat begins withs[0]. Then take the remaining susbtring, reverse it and append it to the begi...
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摘要：Well, this problem has a O(n^3) solution similar to 3Sum. That is, fix two elements nums[i] and nums[j] (i > fourSum(vector& nums, int target) { 2 ...
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摘要：This problem is very similar to 3Sum. You only need to maintain a variable for the sum that is closet to target. Also, some corner cases need to be ha...
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摘要：This is an extension of the 2Sum problem. The idea is pretty simple (even no need to use hash). Sort the array and then starting from the first elemen...
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摘要：Well, it seems that many people meet the TLE problem. Well, I use a simple trick in my code to aoivd TLE. That is, each time before I try to breaks, I...
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摘要：Well, an extension of Remove Duplicates from Sorted Array.The program is fairly similar to that in this solution.1 int removeDuplicates(vector& nu...
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摘要：Well, a medium problem. Use two pointers. One points to the curren number and the other points to the last unique number. Once duplicates occur, move ...
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摘要：Radix sort is another linear time sorting algorithm. It sorts (using another sorting subroutine) the numbers from their least significant digits to mo...
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摘要：Well, this problem is designed for radix sort. For more information about radix sort, Introduction to Algorithms, 3rd edition has some nice examples.H...
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摘要：Counting sort is a linear time sorting algorithm. It is used when all the numbers fall in a fixed range. Then it counts the appearances of each number...
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摘要：The hints on below the problem have suggested a two-pass solution.Now I will focus on the one-pass solution.The one-pass solution has no mystery. Just...
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摘要：This problem gets much trickier than Contains Duplicate and Contains Duplicate II.The basic idea is to maintain a window of k numbers. For each new nu...
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摘要：A classic problem of hash set. The unordered_set of C++ is very suitable for this problem.The code is as follows and it should be quite self-explanato...
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摘要：Problem Description:Given an array of integers that is alreadysorted in ascending order, find two numbers such that they add up to a specific target n...
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摘要：Again, a classic interview question of linked list. Similar to Linked List Cycle, we maintain two pointers fast and slow: fastmove two steps at one ti...
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摘要：A classic interview question of linked list. The idea is to maintain two pointers, one move one step at a time and the other move two steps at a time....
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摘要：This problem has a long story. There are just too many solutions on the web and it can be studied for a long time before you fully grasp it. Morever, ...
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摘要：This is a classic problem for hash table. The basic idea is to maintain a hash table for each element innums, using the element as key and its index (...
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摘要：Well, this problem looks easy at first glance. However, to get a bug-free code may be not that easy.The total square is simply equal to the sum of the...
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摘要：Well, there many ways to solve this problem. Let's first look at a naive solution.The basic idea is simple. Starting from the first character of the s...
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摘要：The key of this problem is that we need not build the tree from scratch. In fact, we can direct obtain its post-order traversal results in a recursive...
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摘要：This is a application of the Trie data structure, with minor extension. The critical part in this problem is to count all the words that have a partic...
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摘要：A direct applicatin of the heap data structure. Specifically, a max heap is used. The required functions include insertion of a node to the heap and e...
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摘要：The idea is to usetwo stacks.Forpush, the first stack records the pushed elements and the second stack recordsallthe minimum (including duplicates) th...
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摘要：This problem is an application of the Trie data structure. In the following, it is assumed that you have solvedImplement Trie (Prefix Tree).Now, let's...
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摘要：Well, the problem does not aim for an advanced algorithm like KMP but only a clean brute-force algorithm. So we can traverse all the possible starting...
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摘要：In this problem, we are asked to divide two integers. However, we are not allowed to use division, multiplication and mod operations. So, what else ca...
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摘要：To readncharacters, we first callread4forn / 4times. For example, if we want to read10characters, we will read them in the8 (4 * 2) + 2manner by first...
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摘要：Problem Description:Design and implement a TwoSum class. It should support the following operations:addandfind.add- Add the number to an internal data...
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摘要：Recently I systematicall review some sorting algorithms, including insertion sort, bubble sort, merge sort and quick sort. I then implement them in C+...
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摘要：Recently I reviewed the classic heapsort algorithm and implement it according to contents in Introduction to Algorithms (3rd edition). The heap data s...
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摘要：The problem has a nice structure that backtracking naturally fits in. The structure is, given a starting positionidx, we search fromidxtill the end of...
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摘要：This problem is somewhat tricky at first glance. However, the final implementation is fairly simple using recursion.The basic idea is, visiting every ...
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摘要：Well, this problem desires for the use of dynamic programming. They key to any DP problem is to come up with the state equation. In this problem, we d...
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摘要：Recursive (Backtracking)This is a typical problem that can be tackled by backtracking.Since backtracking has a more-or-less similar template, so I do ...
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摘要：This problem is similar toBest Time to Buy and Sell Stock. Givenprices, we find the day (denoted asbuy) of the first local minimum and the day (denote...
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摘要：This post shares a very simple solution, whose code is rewritten below, just 5 lines :-) 1 class Solution { 2 public: 3 int maxProfit(vector& pric...
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摘要：This problem seems to be tricky at first glance. However, if you know Morris traversal, it is just the preorder case of Morris traversal and the code ...
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摘要：There are many merge-sort solutions at the forum, but very few quicksort solutions. So I post my accepted quicksort solution here.Well, after reading ...
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摘要：This is a typical DP problem. Suppose the minimum path sum of arriving at point(i, j)isS[i][j], then the state equation isS[i][j] = min(S[i - 1][j], S...
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摘要：Well, this problem is similar toUnique Paths. The introduction of obstacles only changes the boundary conditions and make some points unreachable (sim...
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摘要：This is a fundamental DP problem. First of all, let's make some observations.Since the robot can only move right and down, when it arrives at a point,...
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摘要：Well, this problem has a naive solution, which is to sort the array in descending order and return thek-1-th element. However, sorting algorithm gives...
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摘要：To be honest, this problem is designed to let you use stacks. However, I don't. In fact, you only need to keep a flagand switch it between falseandtru...
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摘要：Well, I do not see what this problem is for. The same code of Binary Tree Level Order Traversal can be used here. The only difference is that we shoul...
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摘要：A classic tree traversal problem. I share my two solutions here: BFS and DFS.BFS: 1 vector> levelOrder(TreeNode *root) { 2 vector> levels;...
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摘要：This is a fundamental and yet classic problem. I share my three solutions here:Iterative solution using stack ---O(n)time andO(n)space;Recursive solut...
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摘要：This is a fundamental and yet classic problem. I share my three solutions here:Iterative solution using stack ---O(n)time andO(n)space;Recursive solut...
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摘要：This is a fundamental and yet classic problem. I share my three solutions here:Iterative solution using stack ---O(n)time andO(n)space;Recursive solut...
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摘要：Well, the basic idea is fairly straightforward. We maintain a mappingmpfrom a value innumsto its position (index)i. Each time we meet an unseen value,...
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摘要：This problem is a little tricky at first glance. However, if you have finished theHouse Robber problem, this problem can simply bedecomposed into two ...
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摘要：Well, have you solved thenextPermutationproblem? If so and you have handled the cases of duplicates at that problem, your code can be used in this pro...
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摘要：Well, have you solved thenextPermutationproblem? If so, your code can be used in this problem. The idea is fairly simple:sortnumsin ascending order, a...
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摘要：Well, in fact the problem of next permutation has been studied long ago. From theWikipedia page, in the 14th century, a man named Narayana Pandita giv...
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