class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
return Core(s,new HashSet(wordDict),0);
}
public boolean Core(String s,Set<String> wordDict,int start){
if(start == s.length()){
return true;
}else{
for(int end = start+1;end<=s.length();++end){
if(wordDict.contains(s.substring(start,end)) && Core(s,wordDict,end)){
return true;
}
}
return false;
}
}
}
第一种暴力遍历递归法,超出时间限制
主要还是动态规划类型题
class Solution {
public boolean wordBreak(String s, List<String> wordDict) {
Set<String> wordDictSet = new HashSet(wordDict);
boolean[] dp = new boolean[s.length()+1];
dp[0]= true;
for(int i = 1;i<=s.length();++i){
for(int j = 0;j<i;++j){
if(dp[j] && wordDict.contains(s.substring(j,i))){
dp[i] = true;
break;
}
}
}
return dp[s.length()];
}
}
把一个问题分成许多个子问题。看子问题是否能得到解决。