leetcode 139. Word Break

最新推荐文章于 2024-04-12 15:24:31 发布

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最新推荐文章于 2024-04-12 15:24:31 发布

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原文链接：http://www.cnblogs.com/jamieliu/p/10415319.html

本文深入探讨了一种用于判断字符串是否能通过字典中的一系列单词进行分割的算法。通过使用布尔数组标记每个可能的单词起始位置，算法能够有效地确定字符串是否符合要求。示例展示了如何利用给定的字典和字符串进行实际操作。

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()];
    }
}