【leetcode】Word Break(java)
Word Break：
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".
大致就是判断：是否字典里的几个词可以串联组成给定的String。
解题过程：
算法不太精通，最开始考虑的就是既然要找下去，那每个结果都不能落下，利用递归可以按照string长度不断递增，在前面把存在于字典的删去，然后再在剩余的里面判断是否可以由字典里的词组成，实现如下：
public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
if(s == null || wordDict == null)
return false;
boolean result = false;
int len = s.length();
if(wordDict.contains(s))
return true;
for(int i = len-1; i >= 0; i--){
String word = s.substring(0,i);
String rest = s.substring(i,len);
if(wordDict.contains(word)){
result = result || wordBreak(rest, wordDict);
}
}
return result;
}
}
然后接酱紫了：- -|||
Status: Time Limit Exceeded
Last executed input:
"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab"
["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]
然后就直接没有思路了，上网查询吧。。。
大家都说可以用动态规划DP做，我自己不是很擅长找最优子结构，然后就看了一眼，里面提到可以用一个一维布尔数组来维护，数组t长度为s.length()+1，t[i]表示 0 到 i-1 是否可以用字典组成（这里指数组序号），逐步进行下去就可以完成，结果是t[s.length()]。我的理解就是可以用它来记录状态，逐步进行下去就可以省略了递归，毕竟可以用之前记录的状态，对于动态规划还是欠缺理解啊。我的自己实现如下：
public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
// t[i]==true => 0-(i-1) can be segmented using dictionary
// t[0]=true
boolean t[] = new boolean[s.length()+1];
t[0] = true;
for(int i = 1; i < s.length()+1; i++){
t[i]=false;
if(wordDict.contains(s.substring(0,i))){
t[i] = true;
continue;
}else{
for(int j = 1; j < i; j++){
if(t[j]&&wordDict.contains(s.substring(j,i))){
t[i] = true;
break;
}
}
}
}
return t[s.length()];
}
}
一会要好好去理解下DP了，我的参考网站如下：
http://www.programcreek.com/2012/12/leetcode-solution-word-break/
欢迎调教。。。。（我好污啊