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CDZSC_2015寒假新人(1)——基础 G
DescriptionIgnatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words an...
分类:其他好文   时间:2015-01-23 13:10:17    阅读次数:115
Leetcode#68 Text Justification
原题地址没有复杂的算法,纯粹的模拟题先试探,计算出一行能放几个单词然后计算出单词之间有几个空格,注意,如果空格总长度无法整除空格数,前面的空格长度通通+1最后放单词、放空格,组成一行,加入结果中对于最后一行要特殊处理代码: 1 vector fullJustify(vector &words, in...
分类:其他好文   时间:2015-01-23 12:54:04    阅读次数:171
Leetcode#30 Substring with Concatenation of All Words
原题地址将L中的单词看成一个整体,这道题与Minimun Window String比较类似,都是利用滑动窗口搜索。所以,依次枚举所有S的起始位置i,从i处开始搜索。当然并不需要枚举所有的i,i最多等于L中单词长度-1。比如L中的单词长度为3,那么当枚举过i=0,1,2后,不用再尝试i=3了,因为结...
分类:其他好文   时间:2015-01-22 20:09:58    阅读次数:189
zsc_寒假训练 7
DescriptionIgnatius likes to write words in reverse way. Given a single line of text which is written by Ignatius, you should reverse all the words an...
分类:其他好文   时间:2015-01-21 21:59:44    阅读次数:192
uva 156 Ananagrams
Most crossword puzzle fans are used to anagrams--groups of words with the same letters in different orders--for example OPTS, SPOT, STOP, POTS and POST. Some words however do not have this attribute, ...
分类:其他好文   时间:2015-01-21 18:22:50    阅读次数:218
UVa 10391 Compound Words(复合词)
题意  输出所有输入单词中可以由另两个单词的组成的词 STL set的应用  枚举每个单词的所有可能拆分情况  看拆开的两个单词是否都存在  都存在的就可以输出了 #include using namespace std; string a, b; set s; set::iterator i; int main() { int l; while(cin >> a) s.i...
分类:其他好文   时间:2015-01-21 18:13:27    阅读次数:210
[C++]LeetCode: 113 Word Break II (DP && Backtacking) 求解拆分组合
题目: Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word. Return all such possible sentences. For example, given...
分类:编程语言   时间:2015-01-20 22:22:42    阅读次数:241
【LeetCode】Substring with Concatenation of All Words
You are given a string, S, and a list of words, L, that are all of the same length. Find all starting indices of substring(s) in S that is a concatena...
分类:其他好文   时间:2015-01-20 17:24:35    阅读次数:125
Leetcode#140 Word Break II
原题地址动态规划题令s[i..j]表示下标从i到j的子串,它的所有分割情况用words[i]表示假设s[0..i]的所有分割情况words[i]已知。则s[0..i+1]的分割情况words[i+1] = words[k] + s[k+1..i+1],其中(有三个条件要满足)(1) 0 wordB....
分类:其他好文   时间:2015-01-20 15:06:21    阅读次数:157
[C++]LeetCode: 107 Reverse Words in a String (2014腾讯实习笔试题)
题目: Given an input string, reverse the string word by word. For example, Given s = "the sky is blue", return "blue is sky the". click to show clarification. Clarification: What co...
分类:编程语言   时间:2015-01-18 14:30:02    阅读次数:239
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