LeetCode : 139. Word Break 拆分字符串

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本文探讨了如何确定一个非空字符串是否可以被分割成一个或多个字典中单词的序列。通过深度优先搜索加剪枝的算法，文章详细解释了如何解决这一问题，并提供了具体的代码实现，包括如何避免超时的方法。

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试题
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

代码
深度优先搜索+剪枝

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> dict = new HashSet<>();
        HashSet<Integer> flag = new HashSet<>();
        for(String d : wordDict){
            dict.add(d);
        }
        return split(s,0,dict,flag);
    }
    public boolean split(String s, int start, HashSet<String> dict, HashSet<Integer> flag){
        if(start==s.length()){
            return true;
        }
        if(flag.contains(start)){
            return false;
        }
        for(int i=start+1; i<=s.length(); i++){
            String l = s.substring(start,i);
            if( dict.contains(l) && split(s,i,dict,flag) )
            {
                return true;
            }
        }
        flag.add(start);
        return false;
    }
}

超时代码

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        HashSet<String> dict = new HashSet<>();
        for(String d : wordDict){
            dict.add(d);
        }
        return split(s,dict);
    }
    public boolean split(String s, HashSet<String> dict){
        if(dict.contains(s)){
            return true;
        }
        if(s.length()==1){
            return false;
        }
        for(int i=1; i<s.length(); i++){
            String l = s.substring(0,i);
            String r = s.substring(i,s.length());
            if( (dict.contains(l)||split(l,dict)) && (dict.contains(r)||split(r,dict)) ){
                return true;
            }
        }
        return false;
    }
}