LeetCode 139 Word Break (DFS 分词解存在性判断)

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#leetcode #DFS #分词

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本文详细解析了LeetCode中的单词拆分问题(Word Break)，通过递归深度优先搜索(DFS)的方法来判断给定字符串是否能由字典中的单词完全匹配组成，并附带完整的Java实现代码。

Given a string s and a dictionary of words dict, determine if s can be segmented into a space-separated sequence of one or more dictionary words.

For example, given
s = "leetcode",
dict = ["leet", "code"].

Return true because "leetcode" can be segmented as "leet code".

题目链接：https://leetcode.com/problems/word-break/

题目分析：比Word Break 2简单，这里不需要记录答案，直接判断即可

public class Solution {
    
    public boolean DFS (int pos, List<String>[] dp, boolean[] flag) {
        if (pos == 0) {
            flag[0] = true;
            return true;
        }
        for (int i = 0; i < dp[pos].size(); i ++) {
            DFS(pos - dp[pos].get(i).length(), dp, flag);
            if (flag[0]) {
                return true;
            }
        }
        return false;
    }
    
    public boolean wordBreak(String s, Set<String> wordDict) {
        int len = s.length();
        if (len == 0 || wordDict.size() == 0) {
            return false;
        }
        List<String>[] dp = new ArrayList[len + 1];
        dp[0] = new ArrayList<>();
        for (int st = 0; st < len; st ++) {
            if (dp[st] != null) {
                for (String word: wordDict) {
                    int ed = st + word.length();
                    if (ed <= len) {
                        if(s.substring(st, ed).equals(word) == true) {
                            if (dp[ed] == null) {
                                dp[ed] = new ArrayList<>();
                            }
                            dp[ed].add(word);
                        }
                    }
                }
            }
        }
        if (dp[len] == null) {
            return false;
        }
        boolean[] flag = new boolean[1];
        flag[0] = false;
        return DFS(len, dp, flag);
    }
}