02 2017 档案
Pocket Gems面经prepare: Diamond and Ruby
摘要：类似Ones and Zeroes Dp[i][j][k] = Max(dp[i-5][j][k] + 5*10, dp[i][j-5][k] + 5*5, dp[i][j][k-1] + 1*25) reduce了dimension，其实还有一个维度是前k个格子，直接用倒序省掉了这个维度
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FB面经Prepare: Find Longest Path in a Multi-Tree
摘要：解法就是在子树里面找最大的两个（或一个，如果只有一个子树的话）高度加起来。 对于每一个treenode, 维护它的最高的高度和第二高的高度，经过该点的最大路径就是： 最高高度+第二高高度，然后return 最高高度
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FB面经Prepare: Friends Recommendation
摘要：有个getFriend() API, 让你推荐你的朋友的朋友做你的朋友，当然这个新朋友不能是你原来的老朋友 1 package fb; 2 3 import java.util.*; 4 5 public class ReferFriends { 6 public List recommendation(String name) { 7 List...
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FB面经Prepare: Dot Product
摘要：1. two pointers 2. hashmap 3. 如果没有额外空间，如果一个很大，一个很小，适合scan小的，并且在大的里面做binary search
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FB面经prepare: Count the number of Vector
摘要：分析： 如果是binary search找每个char的上下界，worst case要找n次，时间复杂度O(nlogn) 所以考虑每次比较start point和start point + 2^n位置上的数，假如一样就continue，不一样就在区间里面binary search找上界，这样wors
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Leetcode: The Maze III(Unsolved Lock Problem)
摘要：Each time, first add the direction to the path, and then go with that direction, checking for hole along the way. When hit a wall, try to turn, and go
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Leetcode: The Maze II
摘要：Solution of The Maze: https://discuss.leetcode.com/topic/77471/easy-understanding-java-bfs-solutionSolution of The Maze III: https://discuss.leetcode.
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Leetcode: The Maze(Unsolved locked problem)
摘要：Search in the four possible directions when coming to a stopping point (i.e. a new starting point). Keep track of places that you already started at i
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Leetcode: Max Consecutive Ones II(unsolved locked problem)
摘要：未研究： The idea is to keep a window [l, h] that contains at most k zero The following solution does not handle follow-up, because nums[l] will need to a
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Leetcode: Find Permutation(Unsolve lock problem)
摘要：还没研究： For example, given IDIIDD we start with sorted sequence 1234567Then for each k continuous D starting at index i we need to reverse [i, i+k] port
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G面经： Design Stock Price Display System
摘要：简单说来PriceUpdate就是添加新的（timestamp, price）, Correction是改之前的(timestamp, price), 求实现当前high(), low(), last() LZ是用的Heap + HashMap, 特别问了时间复杂度(我猜到他想考heap的remov
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U面经Prepare: Print Binary Tree With No Two Nodes Share The Same Column
摘要：这道题若能发现inorder traversal each node的顺序其实就是column number递增的顺序，那么就成功了一大半 维护一个global variable，colNum, 做inorder traversal 然后level order 一层一层打印出来
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U面经Prepare: Web Address
摘要：虽然我觉得这道题用set倒着来就可以解决，但是看到一个Trie的做法也很不错 这里TrieNode.val不再是char, 而是String. children array也变成了Map<String, TrieNode> 我的做法：
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FG面经Prepare: BST to Double LinkedList
摘要：Follow up: 第一步完成基础上，把这个new value插入BST中，一边插入一边记录沿途的大于value的root node（即走了该root node的left child），作为inorder successor. 找到inorder successor之后，通过其prev指针找到in
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G面经Prepare: Longest All One Substring
摘要：这是一个简单版本of LC 424 Longest Repeating Character Replacement 又是Window, 又是Two Pointers Window还是采用每次都try to update left使得window valid, 每次都检查最大window
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G面经Prepare: Print Zigzag Matrix
摘要：For instance, give row = 4, col = 5, print matrix in zigzag order like: [1, 8, 9, 16, 17] [2, 7, 10, 15, 18] [3, 6, 11, 14, 19] [4, 5, 12, 13, 20] 1 package GooglePhone; 2 3 import java.util...
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