07 2015 档案
摘要：Well, this problem seems to be a little tricky at first glance. However, the idea (taken from this link) is really simple. Let's take the equation 2*3...
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摘要：Well, you may need to run some examples to have the intuition for the answer since we only require children with higher rating get more candies than t...
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摘要：Well, a dynamic programming problem. Let's first define its statedp[i][j]to be the number of distinct subsequences oft[0..i - 1]ins[0..j - 1]. Then we...
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摘要：Well, the idea is to search from thetop-rightelement and then reduce the range for further searching by comparisons betweentargetand the current eleme...
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摘要：This problem is more or less the same asFind Minimum in Rotated Sorted Array. And one key difference is as stated in the solution tag. That is, due to...
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摘要：As explained in the Solution tag, the key to solving this problem is to use invariants. We set two pointers:lfor the left andrfor the right. One key i...
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摘要：For those who have already solvedSearch in Rotated Sorted Array, this problem can be solved similarly using codes for that problem and simply adding c...
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摘要：This problem is a nice application of binary search. The key lies in how to determine the correct half fortarget. Since the array has been rotated, we...
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摘要：Well, in the case of a linked list instead of an array, the problem becomes easier. We just need to find the node that will be the new head of the lis...
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摘要：This problem, as stated in the problem statement, has a lot of solutions. Since the problem requires us to solve it in O(1) space complexity, I only s...
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摘要：The first answer in this link has a nice illustrative explanation.Suppose the given array is like [nums[0], nums[1], nums[2], nums[3]]. Then the resul...
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摘要：As the note in the problem statement, this problem has a straight-forward O(n)-space solution, which is to generate the inorder traversal results of t...
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摘要：This problem has a nice BFS structure. Let's illustrate it using the examplenums = [2, 3, 1, 1, 4]in the problem statement. We are initially at positi...
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摘要：This problem has a very concise solution in this link, just 4-lines. The code is rewritten below.1 class Solution {2 public:3 bool canJump(vector&...
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摘要：Well, this problem is just a trick. In fact, we cannot really delete the given node, but just delete its next node after copying the data of the next ...
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摘要：Note: If you feel unwilling to read the long codes, just take the idea with you. The codes are unnecessarily long due to the inconvenient handle of ma...
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摘要：Note: The following idea is in fact from the last answer inthis link, which leads to a clean code. I just reorganize it and add some explanations. I h...
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摘要：Well, an interesting problem. If you draw some examples on a white board and try to figure out the regularities, you may have noticed that the key to ...
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摘要：Well, since we need to make a deep copy of the list and nodes in the list have arandom pointer that may point to any node in the list (or NULL), we ne...
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摘要：Problem DescriptionGiven a string, find the length of the longest substring T that contains at most 2 distinct characters.For example, Given s =“eceba...
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摘要：The key to this problem is to identify all the words that can be added to a line and then identify the places that require an additional space (to mak...
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摘要：An interesting problem! But not very easy at first glance. You need to think very clear about how will a keypoint be generated. Since I only learn fro...
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摘要：Well, a follow-up for the problemLowest Common Ancestor of a Binary Search Tree. However, this time you cannot figure out which subtree the given node...
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摘要：Use the strs[0] as the reference string and then compare it with the remaining strings from left to right. Once we find a string with length less than...
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摘要：Well, a typical backtracking problem. The code is as follows. You may walk through it using the example in the problem statement to see how it works. ...
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摘要：This problem has a very easy recursive code as follows. 1 class Solution { 2 public: 3 bool hasPathSum(TreeNode* root, int sum) { 4 if (!r...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param nums: a vector of integers 5 * @return: an integer 6 */ 7 int findMissing(ve...
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摘要：1 /** 2 * Definition for a point. 3 * struct Point { 4 * int x; 5 * int y; 6 * Point() : x(0), y(0) {} 7 * Point(int a, int b) ...
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摘要：This problem has a naive idea, which is to traverse all possible pairs of two points and see how many other points fall in the line determined by them...
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摘要：A simple application of level-order traversal. Just push the last node in each level into the result.The code is as follows. 1 class Solution { 2 publ...
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摘要：The problem becomes more difficult once the binary tree is not perfect. The idea is still similar to use a level-order traversal. Note that we do not ...
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摘要：The idea is similar to a level-order traversal and remember to take full advantages of the prefect binary tree assumption in the problem statement.The...
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摘要：I guess some of you may have noticed that this seemingly simple problem has the lowest acceptance rate among all problems on the LeetCode OJ. Well, so...
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摘要：Well, remember to take advantage of the property of binary search trees, which is,node -> left -> val val right -> val. Moreover, bothpandqwill be t...
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摘要：The suggested solution to this problem has given a clear idea. The tricky part of this problem is to handle all the edge cases carefully and write a c...
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摘要：The idea is not so obvious at first glance. Since you cannot move from a node back to its previous node in a singly linked list, we choose to reverse ...
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摘要：The key to this problem is to store the values in a stack. In the constructor and next, we add all the next smallest nodes into the stack. The followi...
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摘要：The following idea is taken from a book named 《剑指offer》 published in China.Supposen = 271, it then breaks[1, 271]into[1, 71]and[72, 271]. For[72, 271]...
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摘要：This is a classic problem of linked list. You may solve it using two pointers. The tricky part lies in the head pointer may also be the one that is re...
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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：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：1 class Solution { 2 public: 3 /** 4 * @param A: a vector of integers 5 * @return: an integer 6 */ 7 int firstMissingPosit...
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摘要：This link has a nice explanation of the problem. I rewrite the code below. Play with it using some special examples and you will find out how it works...
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摘要：A classic interview question. This link has a nice explanation of the idea using two stacks and its amortized time complexity.I use the same idea in m...
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摘要：递归实现： 1 class Solution { 2 public: 3 /** 4 * @param nums: A list of integers. 5 * @return: A list of unique permutations. 6 */ 7 ...
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摘要：递归实现：class Solution {public: /** * @param nums: A list of integers. * @return: A list of permutations. */ vector > permute(vector nu...
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摘要：I think the following code is self-explanatory enough. We use anunordered_map countsto record the expected times of each word and anotherunordered_map...
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摘要：By far the best solution I have seen is of O(n) time (some solutions claim to be of O(logn)turns out to be O(n)). One of the simplest ideas is to comp...
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摘要：Well, since theheadpointer may also be modified, we create anew_headthat points to it to facilitate the reverse process.For the example list1 -> 2 -> ...
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摘要：Well, since theheadpointer may also be modified, we create anew_headthat points to it to facilitate the reverse process.For the example list1 -> 2 -> ...
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摘要：Well, since theheadpointer may also been modified, we create anew_headthat points to it to facilitate the swapping process.For the example list1 -> 2 ...
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摘要：动态规划： 1 class Solution { 2 public: 3 /** 4 * @param s: A string 5 * @param p: A string includes "?" and "*" 6 * @return: A boolean...
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摘要：Well, so many people has tried to solve this problem using DP. And almost all of them get TLE (if you see aC++DP solution that gets accepted, please l...
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摘要：Well, the idea of this problem is actually very sample --- keep merging the unmerged lists in lists until there is exactly one list remained. However,...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param s: A string 5 * @param p: A string includes "." and "*" 6 * @return: A boolean 7 ...
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摘要：This problem has a typical solution using Dynamic Programming. We define the state P[i][j] to be true if s[0..i) matches p[0..j) and false otherwise. ...
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摘要：To be honest, I do not know whether this problem is designed to let you use stacks. Anyway, I don't. Here are my codes, both BFS and DFS version. 1 ...
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摘要：1 class Solution { 2 public: 3 /** 4 * @param A: An integer array. 5 * @param B: An integer array. 6 * @return: a double whose for...
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摘要：This link has a very concise and fast solution based on binary search. Spend some time reading it and make sure you understand it. It is very helpful....
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摘要：This is a tough problem! I read some solutions from the web. This link provides a one-end BFS solution. The code is reorganized as follows. 1 class So...
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摘要：This is a classic problem of Dynamic Programming. We define the statedp[i][j]to be the minimum number of operations to convertword1[0..i - 1]toword2[0...
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摘要：The idea to solve this problem is to first sort the intervals according to theirstartfield and then scan the intervals from head to tail and merge tho...
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摘要：A classic subroutine of merge sort. Just merge the elements from back to forth. Keep a pointer for the merged position of the element and two other po...
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摘要：The typical solution to this problem is to use Dynamic Programming. The state dp[i] represents whether s[0..i - 1] can be broken into words in wordDic...
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摘要：This link suggests a concise C++ recursive solution. The original code may be hard to understand at first and I have rewritten the code below. You may...
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摘要：If you have solved the N-Queens problem, this one can be solved in a similar manner. Starting from the first row, we try each of its columns. If there...
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摘要：The idea is to use backtracking. In fact, the code below uses DFS, which involves backtracking in a recursive manner.The idea is also very simple. Sta...
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摘要：1 class Solution { 2 public: 3 /* 4 * @param n: An integer 5 * @return: True or false 6 */ 7 bool checkPowerOf2(int n) { 8 ...
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