随笔分类 - HashMap/ 数组键值对
摘要：Brute Force的做法，N个点两两可以构成N(N-1)/2条线，我们可以找这N(N-1)/2条线中线上点数最大值，只需对每一条线再进行一层O(N)的遍历，总共是O(N^3)。 用第二种方法更好，选一个基准点， 看后面每一个点跟它构成的直线， 维护一个HashMap， key是跟这个点构成直线的
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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：How do we solve problem like this if we were given a normal tree? Yes, traverse it, keep a root to leaf running sum. If we see a leaf node (node.left
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摘要：Taken from GeeksforGeeks Following is a simple algorithm to find out whether a given graph is Birpartite or not using Breadth First Search (BFS) :- Al
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摘要：Analysis: Solution 1: The greedy algorithm is that in each step, select the char with highest remaining count if possible (if it is not in the waiting
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摘要：Design and implement a TwoSum class. It should support the following operations: add and find. add - Add the number to an internal data structure.find
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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：This is a follow up of Shortest Word Distance. The only difference is now you are given the list of words and your method will be called repeatedly ma
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摘要：要注意一种情况：isUnique("cake"), 应该是true. 虽然他的缩写在afterabbr出现，但是那就是他自己，没有别的跟他一样
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摘要：注意这是一一映射，也就是说如果a->dog, b->dog，就应该return false，所以应该在HashMap基础上再加一层检查，即若不含该key，加入map之前应该检查map.values().contains(String) 第二遍：use HashMap和HashSet 用 map.va
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摘要：这道题跟Lintcode: Majority Number II思路很像，那个找大于1/3的，最多有两个candidate，这个一样，找大于1/k的，最多有k-1个candidate 维护k-1个candidate 在map里面，key为数字值，value为出现次数。先找到这k-1个candidat
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摘要：这个是用一个hashtable，key是数字，value是出现次数 然后遍历原数组，每一个数字都把hash里从自己开始往后5个color数都-1，如果发现缺数则说明不能分割 很容易错！ 错了好多次，是color往后5个，如果不存在该color或者color数目已经为0，报错
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摘要：follow up: 建立一个Lookup table, 算过的数就不算了
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摘要：Given n points in the plane that are all pairwise distinct, a "boomerang" is a tuple of points (i, j, k) such that the distance between i and j equals
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摘要：数组;--键值对 follow up:
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摘要：Example 2: Example 3: The logic is very similar to NO.347 and here we just use a map a count and according to the frequency to put it into the right b
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摘要：数组和字符串的分治法: 对返回值的操作(叶结点的+ 非叶结点的), 出口, 分治起始位置, 题目要求--返回值 难点在于从哪分(题目要求的理解? 中间?符号?)和叶结点情况的处理->返回值 见Different Ways to Add Parentheses 投票法: 根据majority 的特性
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摘要：fb: 只用返回true or false。第二题先用set做，后来让改用constant space， 就用了sliding window这样 subarray sum 问题常用hashmap, 存count 值和坐标, 动归的感觉啊 fb:问了数组包含／不包含负数两种情况， 要用 preSum.
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摘要：这种考subarray sum 常用到累加和数组啊, 要清楚从哪到哪开始求和, 在看题意怎么判断就ok了, 数组subarray sum 问题常常用累加和基础上改变 没想到用hashmap O(n) 即可, 关键是存的都是k的余数, 然后余数相见等于零即可 if (k != 0) runningSu
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