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p***n
发帖数: 17190 | 3 玩這個沒啥意思啊
皆 因 習 也
http://www.chinesewords.org/dict/70700-460.html
相沿成習，沿襲。
《魏書·釋老志》：“終恐因習滋甚，有虧恒式。” 唐 崔令欽 《教坊記》：“右多
善歌，左多工舞，蓋相因習。” 清 吳敏樹 《湖北按察使貴陽唐子方先生哀辭》：“
蓋天下數十年來，一切因習頽壞，人材遂靡然不可振發
還有
宋·王安石《答司馬諫議書》：「人習于苟且非一日，士大夫多以不恤國事，同俗自
媚于眾為善。 |
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s**********t
发帖数: 1846 | 5 http://pinyin.sogou.com/dict/news.php?id=421
1975年出生的陈立，前不久在媒人撮合下和与自己同岁的姜玲结婚，他性情羞涩、内向
，“要是我变成了真正的女人，就可以一直被她照顾下去。而且，我也很享受那种被人
照顾的感觉。”
妻子姜玲想变性的理由简单直白，“我要是可以通过手术变成男的，他变成女的。
我再把他娶了，这样婆婆就再也不会欺负我了…我体力上、精神上都比陈立强，所以我
适合变成男的来照顾他。” 姜玲表情坚定，陈立一直在一旁低头羞涩地微笑。他们希
望生下一个孩子再去实施手术。 |
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p*********f
发帖数: 633 | 6 高原之狼国语教学时间：
殴：注音 ㄡ
击、打。如“斗殴”、“殴打”。《红楼梦》第四回：“两家争买一婢，各不相
让，以致殴伤人命”。
(来源：中华民国教育部重编国语辞典 http://dict.revised.moe.edu.tw/)
所以是“……亮国旗险遭殴”，不是“亮国旗险遭欧”。 |
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m*********a
发帖数: 3299 | 7 so in your dict, cost is equal to net income. Please give me a break. You do
not have to preach me anything. You are not a teacher and not qualified to
preach what is cash flow. I do read 10K and 10Q of several of insurancce
companies to invest. |
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m*********a
发帖数: 3299 | 8 The world is not white/back. It is all depending on the degree of regulation
. In my dict, 100% regulation communism, bad. 0% regulation, completely free
market, bad as well.
policy,
greenspan |
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r******y
发帖数: 12263 | 13 clark需要翻译 我说poor and happy boy
clerk又去网上复查, urban dict给的翻译是
Diaosi
28 up, 1 down
A group of people living in the lowest level of city food chain, dreaming
about marrying a stunning beautiful woman (so called goddess), and being
critic for anything that government is going to do, anything. This word is
first invented in one of China's most famous BBS, Baidu tieba. People often
say diaosi is a style of living.
the most famous Diaosi from mainland China is Yi Li, a former soccer player.
buy diaosi... 阅读全帖 |
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f****y
发帖数: 737 | 16 请大家想一想，英语是谁发明的？英国人呗！
英国人认不认识汉语？不认识！
那么英国人在学英语单词的时候需不需要记住单词的汉语意思？不需要，英国人的英语
课本里根本就没有汉字，何谈记住单词的汉语意思？
那么既然英国人学英语不需要记住(甚至根本就见不到)单词的汉语意思，那么中国人学
英语为什么要去记住单词的汉语意思呢？这种做法大家不觉得奇怪吗？
然而由于中国人学英语时都在背单词的汉语意思，因此大家反而觉不出“背汉字”有什
么奇怪的了。
其实仔细想一想，这个行为真的很奇怪，奇怪的根源不在于行为本身，而在于中国人普
遍不会直接识别英语单词的意思，因而只好靠汉语符号来机械地帮助记忆英语单词的意
思，这样去学英语不仅多此一举，而且必然会陷入苦海无边的符号记忆灾难中。
其实英语单词和汉字一样，存在着很多的“偏旁部首”，知道了偏旁部首你就可以根据
它们直接来猜测单词的意思，虽不说百分之百猜准，但起码可以猜测个大概，至少在别
人告诉过你单词的意思后你可以恍然大悟地领会它，这样就可以大大增强你对英语单词
“见字识意”的能力，做到真正认识一个单词，而把它的汉语意思仅做为一般参考。
举几个例子来说吧：
比如单词... 阅读全帖 |
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c******o
发帖数: 1277 | 17 get ShortestUniquePrefix compare a word to a dict
for example
cat against {dog, be, cut} is ca
cat against {dog, be, cut, car} is cat
cat against {dog, be, cut, car, cat} is null
write code, and give running time |
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d******u
发帖数: 1142 | 18 我不大明白为什么要哈希呢？
如果直接用input的字符和dict里面的字符逐个比较，
复杂度不也是 O(h*n)么？
谢谢！ |
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m*****g
发帖数: 226 | 19 1. Modify binary search. when u don't find one, find one. when u find one,
check if it is already first occurrence. if not, look for left half.
2. sounds like unix command?
3. sounds open question?
4. for a lot of digits, the possible permutations may be big
maybe preprocessing dict, convert words into numbers, then search for hit in
the orig digits? |
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V*****i
发帖数: 92 | 20 For the second problem, it is correct to use DP.
But I think the correct formula should be T(n) = \sum_{i=1}^(n-1)(T(i)*H(i+1
, n)), here T(n) the number of partitions for string (1,n), H(i+1, n) if 1
of string (i+1, n) is in dict and 0 if string (i+1, n) is not.
binary
worse |
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z***e
发帖数: 5393 | 24 这有点像我刚听到这个问题时的解法，但是浪费了memory.
Dictionary> dict ----如果一个用户在day 1 visit了page1,然后
day2又visit了page1,那么你在对day1和day2的两个set里面都要加入page1，这就是不
必要的。 |
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z****o
发帖数: 78 | 25 1. use an integer Mark[i] to represent the character set for word[i] in the
dict; use lower 26 bits, each represent whether a certain character appears
in the word.
(Mark[i] & Mark[j]) == 0 means word[i] and
word[j] are disjoint.
2. at the mean time of calculating the Mark[i], we update the maximum-length
word index in
Longest[(~Mark[i]) & 0x03ffffff];
this means eg.
if word[i] = "abcdefg........uvwabcdefg.....uvw", Longest[(~
Mark[... 阅读全帖 |
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l***i
发帖数: 1309 | 26 How much time/space you would spend in building a trie, and update after you
insert you new word? The problem is not explicit about what data structures
are used to store the dict, or what operations it supports with what time
complexity. |
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i**********e
发帖数: 1145 | 27 我写的 boggle 游戏算法，DFS + trie.
一秒以内给出所有 5x5 的答案。
#include
#include
#include
#include
#include
#include
#include
using namespace std;
struct Trie {
bool end;
Trie *children[26];
Trie() {
end = false;
memset(children, NULL, sizeof(children));
}
void insert(const char *word) {
const char *s = word;
Trie *p = this;
while (*s) {
int j = *s-'A';
assert(0 <= j && j < 26);
if (!p->childre... 阅读全帖 |
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k****n
发帖数: 369 | 28 by adv/disadv, you must specify something to compare with...
Then the answer should be trivial, like
Compared with BST, hey can both used to abstract "dictionary". Hash
table is averagely faster, easy to implement, worse worst case
complexity, space waste (actually it is hard to tell)
BST is not only dict, but an ordered set, hash table is not ordered...
Compared with array, linked list, ... ...
You will have more than an interview's time to cover as long as he allows... |
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d********w
发帖数: 363 | 29 拼写检查做法有个计算编辑距离，http://en.wikipedia.org/wiki/Levenshtein_distance
google人工智能老大曾经写过一个非常简短但很高效的拼写检查器，http://norvig.com/spell-correct.html.
下面这个题目也有些类似
Write a program that reads a large list of English words (e.g. from /usr/
share/dict/words on a unix system) into memory, and then reads words from
stdin, and prints either the best spelling suggestion, or "NO SUGGESTION" if
no suggestion can be found. The program should print ">" as a prompt before
reading each word, and should loop until killed.
Yo... 阅读全帖 |
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s*******n
发帖数: 499 | 30 问题是找来的dict big.txt里面，HELLO WORLD没出现过，HELLO也只有两次 |
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g**********y
发帖数: 14569 | 31 public class Boggle {
private final int N = 5;
private HashSet m_words;
public Boggle() {
m_words = new HashSet();
}
public static void main(String[] args) {
String[] str = new String[]{
"AEBOF", "TSUVW", "RFOEG", "RSOFI", "PQWRE"
};
new Boggle().run(str);
}
public void run(String[] str) {
char[][] cs = new char[N][N];
boolean[][] used = new boolean[N][N];
m_word... 阅读全帖 |
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l***i
发帖数: 1309 | 32 One question, can you use the same word in dict more than once?
if yes then wwwyhx gives correct solution. |
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m*****y
发帖数: 93 | 33 one cent：了解python和其他语言不一样的地方，比如内置的list/tuple/dict/set，
yield和generator，list comprehension, lambda function, etc. |
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t********e
发帖数: 143 | 34
Nice. This is the trap. It is not important that characters are swapped.
Just compare them as anagram.
int minSkip(string &s, int start)
{
if (start == s.size())
return 0;
if (m[start] != -1)
return m[start];
//skip first char;
int min = minSkip(s, start+1) + 1;
//used first char case.
for (int i=start; i < s.size(); i++)
{
string sub = s.substr(start, i-start+1);
//use anagram compare for dict lookup
if (isInDict(sub))... 阅读全帖 |
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z*********n
发帖数: 203 | 35 递归比较简单：
string addSpaces(string input, unordered_set &dict) |
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z*********n
发帖数: 203 | 36 递归比较简单：
string addSpaces(string input, unordered_set &dict) |
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l*******8
发帖数: 59 | 37 O(n), n = number of words in dict
★ 发自iPhone App: ChineseWeb 7.7 |
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e****e
发帖数: 418 | 39 Just my 2 cents.
1. Build an interval tree. for each event, check if its start and end are in
direct parent-child relationship. If yes, no conflict for this event.
2. Use a java.util.TreeSet.
for each entry in dict
treeSet.add(entry.firstCharacter).
print out treeSet. |
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k****r
发帖数: 807 | 40 1. A file contains a billion integers, try to find any one integer that is
not in the file.
2. If you wanted to make a highly concurrent cache with a least recently
used replacement policy, what data structures would you use? How would this
scale per number of threads?
3. How would you implement hash table on your own? Write the code for
implementing your own hash table?
4. Generate a random 4 letter word from /usr/share/dict/words |
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e***s
发帖数: 799 | 42 二爷，是要把分割的单词返回到一个字符串中，方法定义是
String segmentString(String s, Set dict) |
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w***o
发帖数: 109 | 43 这样行不行？只能找到最少的个数，加个数组，稍改一下就能找到如何分割。
int n = s.length();
int[] dp = new int[n+1];
Arrays.fill(dp, 1000000);
dp[0] = 0;
for(int i = 1; i <= n; i++) {
for(int j = i-1; j >= 0; j--) {
if(dict.contains(s.substring(j, i)))
dp[i] = Math.min(dp[i], dp[j]+1);
}
}
if(dp[n] > n)
System.out.println(“Not valid”);
else
System.out.println(“” + dp[n]); |
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d*********g
发帖数: 154 | 44 for(int i = 0; i < vs.size(); i++)
{
if(findLongestWord_helper(vs[i], hashmap))
{
result = vs[i];
break;
}
}
为什么这里break了？不是要找最长的么？你这里只是找到一个由dict里面的单词构成
的单词之后就break了吧？ |
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B********t
发帖数: 147 | 45 我把dict sort了的 从最长的开始找 找到就可以退出了 |
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c********t
发帖数: 5706 | 46 好像弄明白你的题意了。虽然我不懂c++,不过我觉得你的问题是，findLongestWord_
helper又想split to several words才能true, 又想如果剩一个词也要true
看看改成下面的行不
bool findLongestWord_helper(string &s, map &hm，bool hasSplit)
{
for(int i = 1; i <= s.size(); i++)
{
string sub_f(s, 0, i);
if(hm[sub_f])
{
if(i==s.size() ) return hasSplit;
string sub_b(s, i, s.size()-i);
if(findLongestWord_helper(sub_b, hm, true))
return true;
}
}
return... 阅读全帖 |
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w****a
发帖数: 710 | 47 "10分钟前面经"系列的第三弹，这次是F家电二面。
不用大家Bless了，这次必挂了。就问了一道技术题。
Word breaking
// How many spaces can we add to a word such that:
// All subwords can be found within a given dictionary
// fireman
// fire man -> 2 words
// fir em an -> 3 words
/* DICT
a
an
em
fir
fire
ire
ma
man
*/
这题以前没做过，我当时想这个看上去就是个典型DP题。
但是没做过不敢妄为 就想先用回溯先做个解出来再跟他讨论DP。
然后回溯做法出了个bug被他指出来，然后卡壳了好一会。最后DP也没做成。面试时间
一共三十分钟左右。就因为卡了会壳估计面试官都没兴趣让我再用别的解法做了。。
而且中途还有个，我循环用i和j被他指出来了。我正常变量命名都还行，就是循环我一
直喜欢用i，以后连循环我都不敢用i了。大家吸取教训。
给大家个教训吧。没做过的题千万别慌。写出代码了自己一... 阅读全帖 |
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w****x
发帖数: 2483 | 48
感觉正常的思路是有一步pre processing,就是先把dict建图，
然后后续的查询ladder从建好的图为出发点。
这个图可以用map表示 |
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y****e
发帖数: 1012 | 49 DFS+pruning?
那步dict.erase()用的很妙
展开说说怎么想到的吧~
vector< |
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w****x
发帖数: 2483 | 50
感觉正常的思路是有一步pre processing,就是先把dict建图，
然后后续的查询ladder从建好的图为出发点。
这个图可以用map表示 |
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