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()];
    }
}

leetcode 139. Word Break

标签：pac   rdd   nat   OLE   can   output   sum   一个   nbsp   

原文地址：https://www.cnblogs.com/jamieliu/p/10415319.html