Word Break - Leetcode

public class Solution {
public boolean wordBreak(String s, Set<String> dict) {
boolean[] f = new boolean[s.length()+1];
f[0]=true;
for(int i=1; i<=s.length(); i++){
for(int j=i-1; j>=0; j--){
String sub = s.substring(j, i);
if(f[j]==true && dict.contains(sub)){
// dict.remove(sub);
f[i] = true;
break;
}
}
}
return f[s.length()];
}
}

hint: dict里面的单词可以重复出现；如果i-j 属于dict, 这时候前提是0-i 属于dict 则0-j才是dict的子元素。

思想： 动规。 之前的值会影响后面的值。

O(n^2) / O(n)

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"

.