Leetcode-139. Word Break

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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"

.
这个题目提醒了我，我一开始用Set只保存了对的字符串，结果超时了，我看了别人的解法是保存错的字符串。。想了想确实，因为错的比对的情况多很多。Your runtime beats 5.13% of java
submissions
public class Solution {
private Set<String> invalidStr = new HashSet<String>();
public boolean wordBreak(String s, Set<String> wordDict) {

boolean flag = false;

if(wordDict.contains(s)) return true;
if(invalidStr.contains(s)) return false;

for(int i = 1; i < s.length(); i ++){
String subString = s.substring(0,i);
if(wordDict.contains(subString)) flag = wordBreak(s.substring(i,s.length()),wordDict);
if(!flag)invalidStr.add(s);
else break;
}
return flag;
}
}

后来我想了想，我也可以把对的保存了。。不过我这样还是有点麻烦，看了下dp的思路，其实我这个也算是dp，用一个dp的array保存哪些位置是true确实更好一些。。Your runtime beats
16.34% of java submissions.
public class Solution {
private Set<String> invalidStr = new HashSet<String>();
private Set<String> validStr = new HashSet<String>();
public boolean wordBreak(String s, Set<String> wordDict) {

boolean flag = false;

if(wordDict.contains(s) || validStr.contains(s)) return true;
if(invalidStr.contains(s)) return false;

for(int i = 1; i < s.length(); i ++){
String subString = s.substring(0,i);
if(wordDict.contains(subString)) flag = wordBreak(s.substring(i,s.length()),wordDict);
if(!flag)invalidStr.add(s);
else{validStr.add(s); break;}
}
return flag;
}
}