139. 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"

.

分析：动态规划；

设置一个boolean数组，其中每个元素用来代表字符串s的前i个字符串，是不是能够用set中的字符串表示。

时间复杂度：O(N*N)

空间复杂度：O(N)

public class Solution {
public boolean wordBreak(String s, Set<String> wordDict) {
int n=s.length();
boolean[] flag=new boolean[n+1];
flag[0]=true;
for(int i=1;i<=n;i++){
for(int j=0;j<=i;j++){
String temp=s.substring(j,i);
if(flag[j]&&wordDict.contains(temp)){
flag[i]=true;
}
}
}
return flag
;
}
}