leetcode12.wordBreak

代码如下：

package leetcode10.WordBreak;
import java.util.HashSet;
import java.util.LinkedHashSet;
import java.util.Set;

/**
Word Break 
 
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".

 *
 */
public class S147 {

	public static void main(String[] args) {
		Set<String> dict = new HashSet<String>();
		dict.add("aaa");
		dict.add("aaaa");
		String s = "aaaaaaa";
		System.out.println(wordBreak(s, dict));
		System.out.println(wordBreak2(s, dict));
	}

	// DP
	public static boolean wordBreak(String s, Set<String> dict) {
		// 如果canBreak[i]为true，则s[0...(i-1)]能被拆分
		boolean[] canBreak = new boolean[s.length()+10];
		canBreak[0] = true;		// 初始化
		
		for(int i=0; i<s.length(); i++){
			if(canBreak[i] == false){
				continue;
			}
			for(String dictS : dict){
				int len = dictS.length();
				int end = i + len;
				if(end > s.length()){
					continue;
				}
				if(s.substring(i, end).equals(dictS)){
					canBreak[end] = true;
				}
			}
		}
		return canBreak[s.length()];
	}
	
	// DFS TLE
	public static boolean wordBreak2(String s, Set<String> dict) {
		return dfs(s, dict, 0);
    }
	
	public static boolean dfs(String s, Set<String> dict, int start){
		if(start >= s.length()){
			return true;
		}
		
		for(String dictS : dict){
			int len = dictS.length();
			if(start+len<=s.length() && s.substring(start, start+len).equals(dictS)){
				if(dfs(s, dict, start+len)){
					return true;
				}
			}
		}
		
		return false;
	}
	
}

总结：DFS的方法要掌握，思路可以理解为。从字典中拿出一个字段与字符串的子串对比，如果符合，剩下的利用DFS递归结果也为真，结果即为真，递归的思想

原文地址：http://blog.youkuaiyun.com/fightforyourdream/article/details/17498349