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Monday, November 18, 2013

[LeetCode] Word Break, Solution

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

[Thoughts]

一个DP问题。定义possible[i] 为S字符串上[0,i]的子串是否可以被segmented by dictionary.

那么

possible[i] = true      if  S[0,i]在dictionary里面

                = true      if   possible[k] == true 并且 S[k+1,j]在dictionary里面, 0<k<i

               = false      if    no such k exist.

 

[Code]

实现时前面加一个dummy节点,这样可以把三种情况统一到一个表达式里面。

1 bool wordBreak(string s, unordered_set<string> &dict) {
2 string s2 = '#' + s;
3 int len = s2.size();
4 vector<bool> possible(len, 0);
5
6 possible[0] = true;
7 for(int i =1; i< len; ++i)
8 {
9 for(int k=0; k<i; ++k)
10 {
11 possible[i] = possible[k] &&
12 dict.find(s2.substr(k+1, i-k)) != dict.end();
13 if(possible[i]) break;
14 }
15 }
16
17 return possible[len-1];
18 }
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