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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.
Given
s = "lintcode",
dict = ["lint", "code"].
Return true because "lintcode" can be segmented as "lint code".
Analysis:
It is a DP problem. However, we need to use charAt() instead of substring() to optimize speed. Also, we can first check whether each char in s has appeared in dict, if not, then directly return false. (This is used to pass the last test case in LintCode).
Solution:
1 public class Solution { 2 /** 3 * @param s: A string s 4 * @param dict: A dictionary of words dict 5 */ 6 public boolean wordSegmentation(String s, Set<String> dict) { 7 if (s.length()==0) return true; 8 9 char[] chars = new char[256]; 10 for (String word : dict) 11 for (int i=0;i<word.length();i++) 12 chars[word.charAt(i)]++; 13 14 for (int i = 0;i<s.length();i++) 15 if (chars[s.charAt(i)]==0) return false; 16 17 boolean[] d = new boolean[s.length()+1]; 18 Arrays.fill(d,false); 19 d[0] = true; 20 for (int i=1;i<=s.length();i++){ 21 StringBuilder builder = new StringBuilder(); 22 for (int j=i-1;j>=0;j--){ 23 builder.insert(0,s.charAt(j)); 24 String cur = builder.toString(); 25 if (d[j] && dict.contains(cur)){ 26 d[i]=true; 27 break; 28 } 29 } 30 } 31 32 return d[s.length()]; 33 34 } 35 }
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原文地址：http://www.cnblogs.com/lishiblog/p/4183748.html
