【题目】
Given a string S and a string T, count the number of distinct subsequences of T in S.
A subsequence of a string is a new string which is formed from the original string by deleting some (can be none) of the characters without disturbing the relati...
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其他好文 时间:
2014-06-29 07:27:17
阅读次数:
210
http://unicode.org/reports/tr35/tr35-6.html#Date_Format_PatternsAppendix F:Date
Format PatternsA date pattern is a string of characters, where specifi...
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其他好文 时间:
2014-06-11 23:37:19
阅读次数:
499
1.1Implement an algorithm to determine if a
string has all unique characters. What if you cannot use additional data
structures?字符串问题,需要先确定是不是只有ASCII码...
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其他好文 时间:
2014-06-11 12:27:53
阅读次数:
247
Given a string, find the length of the longest substring without repeating characters. For example, the longest substring without
repeating letters for "abcabcbb" is "abc", which the length is 3. Fo...
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其他好文 时间:
2014-06-08 16:27:47
阅读次数:
231
Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified.
You should pack your words in a greedy approach; that i...
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编程语言 时间:
2014-06-08 15:25:00
阅读次数:
282
字符串和字符 (Strings and Characters)String是一个有序的字符集合,例如
"hello, world", "albatross"。Swift 字符串通过String类型来表示,也可以表示为Character类型值的集合。Swift
的String和Character类型提...
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其他好文 时间:
2014-06-08 00:24:00
阅读次数:
381
【题目】
Given an array of words and a length L, format the text such that each line has exactly L characters and is fully (left and right) justified.
You should pack your words in a greedy approach; that is, pack as many words as you can in each line. Pad ...
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其他好文 时间:
2014-06-04 22:37:46
阅读次数:
389
【题目】原文:1.3 Design an algorithm and write code
to remove the duplicate characters in a string without using any additional
buffer. NOTE: One or two add...
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其他好文 时间:
2014-06-02 21:32:04
阅读次数:
284
题目描述:题目描述相当的繁琐
分析:输入保证填词游戏至少有一组答案,这就说明不必寻找单词所在的位置,只要去掉这些单词所占用的字母就可以了。 代码实现一下: #include void
main(){ int characters[26]; int n, m, p; int i, j; for(i =...
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其他好文 时间:
2014-06-02 08:22:49
阅读次数:
208
题目:
给定一个字符串,返回该串没有重复字符的最长子串。
分析:
1)子串:子串要求是连续的。
2)无重复,出现重复就断了,必须从新的位置开始。而新的位置就是重复字符第一次出现位置的下一个位置。
3)整个串可能没有一处重复。
那么,为了找出当前访问的字符是否出现过,要怎么做呢?当然是hash,O(1)的时间,而且既然是字符, 定义个255的hash table 就可以了,has...
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其他好文 时间:
2014-06-01 10:48:31
阅读次数:
206