Word Break

139. Word Break

Difficulty: Medium

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

Solution

Language: Java

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        // dp[i]表示0-i可以切分
        boolean[] dp = new boolean[s.length()+1];
        //　初始化
        dp[0] = true;
        for(int i=1;i<=s.length();++i){
            for(int j=0;j<i;++j){
                // 如果0-j已经正常切分，并且j-i也是一个单词，那么 0-i可以切分
                if(dp[j] && wordDict.contains(s.substring(j, i))){
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.length()];
        
    }
}