43.WordBreak（看字符串是否由词典中的单词组成）

Level：

  Medium

题目描述：

Given a non-empty string s and a dictionary wordDict containing a list of non-empty words, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

Note:

• The same word in the dictionary may be reused multiple times in the segmentation.


• You may assume the dictionary does not contain duplicate words.

Example 1:

Input: s = "leetcode", wordDict = ["leet", "code"]
Output: true
Explanation: Return true because "leetcode" can be segmented as "leet code".

Example 2:

Input: s = "applepenapple", wordDict = ["apple", "pen"]
Output: true
Explanation: Return true because "applepenapple" can be segmented as "apple pen apple".
Note that you are allowed to reuse a dictionary word.

Example 3:

Input: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"]
Output: false

思路分析：

  题目要求判断给出的字符串是否能由词典中的单词组成，我们应该用动态规划的思想去解决。但凡是能把问题规模缩小的都应该想到用动态规划求解。例如本题，如果我知道给定字符串的0到i子串可以用字典中的单词表达，那么我只需要知道i+1到末尾的子串能否被字典表达即可知道整个字符串能否被字典表达。所以随着i的增大，问题规模逐渐的缩小，且之前求解过的结果可以为接下来的求解提供帮助，这就是动态规划了。设dp[i]代表s.substring(0, i)能否被字典表达，此刻我们知道dp[0]~dp[i-1]的结果。而dp[i]的结果由两部分组成，一部分是dp[j]（j

代码：

public class Solution{
public boolean wordBreak(String s,ListwordDict){
if(s==null||s.length()==0)
return false;
boolean []dp=new boolean [s.length()+1]; //dp[i]表示字符串的前i个字符是否能由词典中的单词表示。
dp[0]=true;
for(int i=1;i<=s.length();i++){
for(int j=0;j if(dp[j]&&wordDict.contains(s.substring(j,i))){
dp[i]=true;
break;
}
}
}
return dp[s.length()];
}
}