leetcode 139. Word Break
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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
用一个boolean数组来标记每一个word的起始位置。当该字母为一个新单词的开头时,该位置被标记为true(包括最后一个单词的下一个位置,即s.length()的位置)。
class Solution { public boolean wordBreak(String s, List<String> wordDict) { boolean[] f = new boolean[s.length()+1]; f[0] = true; for(int i = 1; i <= s.length(); i++) { for(int j = 0; j < i; j++){ if(f[j] && wordDict.contains(s.substring(j,i))){ f[i] = true; break; } } } return f[s.length()]; } }
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