随笔分类 - FB
摘要：Implement a basic calculator to evaluate a simple expression string. The expression string contains only non-negative integers, +, -, *, / operators a
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摘要：Use buckets to record the last position of digit 0 ~ 9 in this num. Loop through the num array from left to right. For each position, we check whether
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摘要：The sample input is very confusing when time t has mixed meaning of beginning of time t for start and end of time t for end We can increase all end ti
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摘要：注意这道题：Must be exactly one distance apart. Not the same. 另外FB面经有一道比较狠的这个题的变形： 这题就是one edit distance的变形题，难点在于给的Iterator，事先不知道两个file的长度，也不允许用extra space（
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摘要：小心这里是index triplets, 不是nums[i]数组元素的triplets, 所以3Sum那道题里面的跳过条件不用了 因为不关心每个index具体是什么，只关心个数，所以可以排序
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摘要：Time Complexity will be O(n) because the "start" and "end" points will only move from left to right once.
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摘要：我又想到了另一个idea，使用那个assumption可以达到O(1) 复杂度：首先判断isUp or not，仅以isUp为例, 假设A[x] 是最高点，则有：A[x] - A[i] = x - i. A[x] - A[j] = j - x由此可以解出 x = 1/ 2 * (A[j] - A[i
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摘要：A->B->C->D->E | V M | V N 有right和down的linkedlist，要求更新linkedlist变成A->M->N->B->C->D->E,注意M可以还有right，B也可以有down，其实就是递归输出。然后我做了一会才做出，之前有个大神的面经里发过，但我没仔细看很后悔。我昨晚发现这tmd不是inorder traversal么 // Class to ...
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摘要：写了一行，面试官打断我，让我再说一下思路，我又说了一下准备定义一个array，n+1的size，第一个存0，保存到当前值的所有之前numbers的和，然后比如query（0，3），就能用preSum[4]-preSum[0];他表示不懂...我又举了3个例子...面试官表示ok，但你不要写代码，先写
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摘要：The idea is to use two arrays len[n] and cnt[n] to record the maximum length of Increasing Subsequence and the coresponding number of these sequence w
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摘要：Write a function to find the longest common prefix string amongst an array of strings.第二遍做法：时间复杂度应该是O(m*n)，m表示字符串的最大长度，n表示字符串的个数，空间复杂度应该是O(m),即字符串的长度 public class Solution { public String longest...
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摘要：Solution 1: Reservior sampling: (wiki introduction) Reservoir sampling is a family of randomized algorithms for randomly choosing a sample of k items
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摘要：Strobogrammatic Number II give us all the strings with length n, then we can call it to get strings with length between low.length() and high.length()
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摘要：参考http://segmentfault.com/a/1190000003787462 找出所有的可能，必然是深度优先搜索。但是每轮搜索如何建立临时的字符串呢？因为数是“对称”的，我们插入一个字母就知道对应位置的另一个字母是什么，所以我们可以从中间插入来建立这个临时的字符串。这样每次从中间插入两个
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摘要：A strobogrammatic number is a number that looks the same when rotated 180 degrees (looked at upside down). Write a function to determine if a number is strobogrammatic. The number is represented as ...
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摘要：The general solution would be to check each char in turn and report the first char that makes the string a non number. There are some exceptions to wa
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摘要：Given a string, your task is to count how many palindromic substrings in this string. The substrings with different start indexes or end indexes are c
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摘要：Given two strings s and t, determine if they are isomorphic. Two strings are isomorphic if the characters in s can be replaced to get t. All occurrenc
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摘要：/*Given a string and dictionary of words, break the string into minimum number of words from the dictionary. Ex: {"jumped", "over", "some", "thing”, "
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摘要：字符串匹配的问题应该最先想到dp, 主要还是说一下动态规划的方法。跟Regular Expression Matching一样，还是维护一个假设我们维护一个布尔数组res[i],代表s的前i个字符和p的前j个字符是否匹配(这里因为每次i的结果只依赖于j-1的结果，所以不需要二维数组，只需要一个一维数
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