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;    }}