題目：Single Number

難度：medium

問題描述：

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

解題思路：

  * 使用動態規劃
  * 使用temp[i]代表s的前i個字元是否能被切割
  * temp[i]=
  *                  true :任意 j<i, （temp[j]=true）&&（s[j,i]屬於wordDict）
  *                  false：其他情況

程式碼如下：

public static boolean wordBreak(String s, List<String> wordDict) {
        if(wordDict.size()==0) return false;
        boolean[] dp = new boolean[s.length()+1];//在該位置之前是否是可行的，也就是該處可以加分隔符
        dp[0] = true;//空字串是true
        for(int i=1;i<dp.length;i++){//求出dp[i] 也就是前i個字元 s[0]-s[i-1]
        	for(int j=i-1;j>=0;j--){ 
        		if(dp[j]){//前j個可以分割 s[0]-s[j-1]
        			String str = s.substring(j,i);
        			if(wordDict.contains(str)){
        				dp[i] = true;
        				break;
        			}
        		}
        	}
        }
        return dp[s.length()];
    }