Leetcode(139) Word Break

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Description

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:

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

Example 2:

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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:

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

解法

动态规划的思想解题，采用数组flag[i]记录字符串到i位置(不含i)是否可分，对于每个位置，遍历分割点然后更新数组即可

具体代码如下：

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class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        int len = s.length();
        boolean[] flag = new boolean[len+1];
        flag[0] = true;
        for(int i = 1; i<len+1; i++){
            for(int j = 0; j < i; j++){
                if(flag[j]==true&&wordDict.contains(s.substring(j,i))){
                    flag[i] = true;
                    break;
                }
            }
        }
        return flag[len];
    }
}