Word Break I,II
Given a string s and a dictionary of words dict, determine ifs 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".
public class Solution { // http://needjobasap.blogspot.com/2014/11/word-break-leetcode.html //http://fisherlei.blogspot.com/2013/11/leetcode-word-break-solution.html public boolean wordBreak(String s, Set<String> dict) { boolean[] dp = new boolean[s.length()+1]; dp[0] = true; // for(int i=0;i<s.length();i++){ for(int i=1;i<=s.length();i++){ for(int j=0; j< i; j++){ if(dp[j] && dict.contains(s.substring(j,i))) { dp[i]=true; break; } } } return dp[s.length()]; } }
II
Given a string s and a dictionary of words dict, add spaces in s to construct a sentence where each word is a valid dictionary word.
Return all such possible sentences.
For example, given
s = "catsanddog",
dict = ["cat", "cats", "and", "sand", "dog"].
A solution is ["cats and dog", "cat sand dog"].
就是在1 的基础上加了个dfs的应用
public class Solution { public ArrayList<String> wordBreak(String s, Set<String> wordDict) { ArrayList<String> res = new ArrayList<String>(); if(s==null||s.length()==0) return res; int len = s.length(); boolean [] dp = new boolean[len+1]; dp[0] = true; for(int i=1;i<=len;i++){ for(int j=0;j<i;j++){ String tmp = s.substring(j,i); if(dp[j] && wordDict.contains(tmp)){ dp[i] = true; break; } } } if(dp[len]==false) return res; StringBuilder item = new StringBuilder(); dfs(s, 0, item, res, wordDict); return res; } private void dfs(String s, int start, StringBuilder item, ArrayList<String> res, Set<String> dict){ int len = s.length(); if(start>=len){ res.add(new String(item)); return; } for(int i=start+1;i<=len;i++){ String sub = s.substring(start,i); if(dict.contains(sub)){ int ol = item.length(); if(ol!=0){ item.append(" "); } item.append(sub); dfs(s,i,item,res,dict); item.delete(ol,item.length()); } } } }