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The idea is to add a set to the hashMap to remember all the locations of a duplicated number.
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posted @ 2016-11-29 02:44
neverlandly
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最先想到是用double LinkedList+Map, 没必要，arraylist+map就够了；另外取random的方法还有，rand.nextInt(int n) returns an integer in the range [0, n) java.util.Random rand = ne
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posted @ 2016-11-29 01:27
neverlandly
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Heap: you need to know the row number and column number of that element(so we can create a tuple class here) Binary Search方法：（12/28仔细看了之后觉得没必要深究，有时间再去
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posted @ 2016-11-28 13:02
neverlandly
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DP 解法： the key to solve DP problem is to think about how to create overlap, how to re-solve subproblems（怎么制造复用） Bottom up dp: Better Solution(Bottom-u
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posted @ 2016-11-28 11:40
neverlandly
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DP solution is O(N^2), better way is greedy Now for explanation, we take example series:2,1,4,5,6,3,3,4,8,4 First we check if the series is starting a
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posted @ 2016-11-28 10:26
neverlandly
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Clarification of the problem: https://discuss.leetcode.com/topic/68252/clarification-on-the-problem-description-problem-description-need-to-be-updated
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posted @ 2016-11-28 06:43
neverlandly
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Binary Search Here "My" means the number which is given for you to guess not the number you put into guess(int num).
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posted @ 2016-11-28 05:33
neverlandly
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Better Solution: O(KlogK), 转自https://discuss.leetcode.com/topic/50885/simple-java-o-klogk-solution-with-explanation Some observations: For every numbe
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posted @ 2016-11-28 04:55
neverlandly
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ab % k = (a%k)(b%k)%kSince the power here is an array, we'd better handle it digit by digit.One observation:a^1234567 % k = (a^1234560 % k) * (a^7 % k
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posted @ 2016-11-28 01:48
neverlandly
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DP Solution similar to Longest Increasing Subsequence: 我的解法：用一个arraylist存以某一个element结尾的最长sequence 别人的好方法，大体思路一样，只是不存arraylist, 而用一个preIndex数组存以某一个elem
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posted @ 2016-11-27 01:24
neverlandly
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参考：https://discuss.leetcode.com/topic/49238/math-solution-java-solution The basic idea is to use the property of Bézout's identity and check if z is a
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posted @ 2016-11-27 00:10
neverlandly
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转自https://discuss.leetcode.com/topic/49771/java-simple-easy-understand-solution-with-explanation/2，注意里面对于减法的讲解 have been confused about bit manipulati
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posted @ 2016-11-26 11:53
neverlandly
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Reference: https://discuss.leetcode.com/topic/48875/accepted-c-codes-with-explanation-and-references/2 The naive solution is brute-force, which is O((
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posted @ 2016-11-26 08:01
neverlandly
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Analysis: A number of unique digits is a number which is a combination of unrepeated digits. So, we can calculate the total number. for number with n
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posted @ 2016-11-26 04:20
neverlandly
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注意需要maintain一个time stamp, 每次postTweet时++。还有要注意unfollow的时候，不能unfollow自己 关于PriorityQueue还看到这一种写法，挺简单的： 1 Set<Integer> users = userMap.get(userId).follow
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posted @ 2016-11-26 01:58
neverlandly
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1-9 : count:9 * len:1 10-99： count:90 * len:2 100-999: count:900 * len:3 1000-9999: count: 9000 * len:4 maintain a count, len, start
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posted @ 2016-11-25 06:51
neverlandly
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DP 解法， Time Complexity: O(N^2) sort based on width or height, anyone is ok. After the sorting, for each envelope, those envelopes which can fit into t
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posted @ 2016-11-25 05:36
neverlandly
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TreeMap 解法： Use TreeMap to easily find the lower and higher keys, the key is the start of the interval.Merge the lower and higher intervals when neces
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posted @ 2016-11-25 02:06
neverlandly
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我的binary Search 解法：无需变成long 别人三种方法总结：
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posted @ 2016-11-24 12:50
neverlandly
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用HashMap Follow Up 1: 用two pointer解，可以省存成HashMap Follow Up 2: 用在长的数组里面binary Search可解 Follow Up 3: What if elements of nums2 are stored on disk, and t
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posted @ 2016-11-24 11:52
neverlandly
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Use two hash sets Time complexity: O(n) 注意Set<Integer> set = new HashSet<>(); 第二个泛型可以不写出 Iterator iter = set2.iterator(); for (int i=0; i<res.length;
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posted @ 2016-11-24 10:34
neverlandly
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很像majority element III, 但是那道题有O(k) extra space的限制，这里没有。有任意extra space, 同时知道elem range情况下，bucket sort最节省时间 语法上注意第5行，建立一个ArrayList的array，后面没有泛型generic L
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posted @ 2016-11-24 06:51
neverlandly
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Two Pointers 解法， 注意大小写
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posted @ 2016-11-24 05:20
neverlandly
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O(N^2)解法： DP dp[i] represent the maximum product of breaking up integer i O(N)解法： the best factor is 3. we keep breaking n into 3's until n gets small
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posted @ 2016-11-24 04:57
neverlandly
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非常精巧地使用stack。push all the nestedList into the stack from back to front，so when we pop the stack, it returns the very first element 执行hasNext()的时候，如果pe
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posted @ 2016-11-24 01:29
neverlandly
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After the API changes 注意没有append(0, s.charAt(i)), 是sb.insert(0, charAt(i));
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posted @ 2016-11-23 12:20
neverlandly
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it's easy to find that power of 4 numbers have those 3 common features. First,greater than 0. Second,only have one '1' bit in their binary notation,so
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posted @ 2016-11-23 11:48
neverlandly
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https://discuss.leetcode.com/topic/39834/step-by-step-tackling-of-the-problem rob(root) which will return the maximum amount of money that we can rob
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posted @ 2016-11-23 11:17
neverlandly
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posted @ 2016-11-23 07:36
neverlandly
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Naive Solution: Time: O(n^2*k) with n the total number of words in the "words" array and k the average length of each word: check each combination see
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posted @ 2016-11-23 06:29
neverlandly
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4th line may cross with 1st line, and so on: 5th with 2nd, ...etc 5th line may cross with 1st line, and so on: 6th with 2nd, ...etc 6th line also may
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posted @ 2016-11-23 03:51
neverlandly
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Naive Solution: use DP, Time O(N^2), Space O(N) dp[i] represents the length of longest increasing subsequence till i including element i in nums array
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posted @ 2016-11-23 01:42
neverlandly
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我的想法：假设给定的点事ListNode oneNode, 设置ListNode dummy = new ListNode(-1); dummy.next = oneNode 然后扫描一次找到环里最后一个点，断链：ListNode end.next = null; 然后就开始扫描，删点 然后再扫描一
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posted @ 2016-10-19 00:49
neverlandly
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设计一个shuffle card 用了java. Random Class 1 package Random; 2 import java.util.*; 3 4 public class Solution { 5 static int cardNum = 10; 6 public int[] sh
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posted @ 2016-02-17 04:50
neverlandly
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The definition of integer average is the highest smaller integer if average is floating point number. Also the condition if that they can not use any
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posted @ 2016-02-16 11:56
neverlandly
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Given an array which has n integers,it has both positive and negative integers.Now you need sort this array in a special way.After that,the negative i
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posted @ 2016-02-16 07:40
neverlandly
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给定一个char array， 这个array是一个句子，然后给定一个字母，把这个array里面带有这个字母开头的单次删掉，操作是要求in place. 检测 array[i]==' ' && i<array.length-1 && array[i+1]==target，这种情况，设置j从i+1开始
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posted @ 2016-02-16 04:15
neverlandly
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You have two every large binary trees: T1, with millions of nodes, and T2, with hundreds of nodes. Create an algorithm to decide if T2 is a subtree of
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posted @ 2016-02-08 23:45
neverlandly
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refer to Recursion https://leetcode.com/discuss/84702/share-my-solution and Iteration https://leetcode.com/discuss/84706/share-solution-java-greedy-st
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posted @ 2016-02-08 12:29
neverlandly
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You are given a binary tree (not necessarily BST) in which each node contains a value. Design an algorithm to print all paths which sum up to that val
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posted @ 2016-02-04 12:02
neverlandly
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