Leetcode: wordBreak

最新推荐文章于 2025-08-11 15:06:51 发布

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#leetcode #动态规划

本文探讨了如何判断一个字符串是否可以被分割成一个或多个字典中存在的单词的问题，并提供了一种递归解决方案。通过示例说明了如何使用该方法进行字符串的分割验证。

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题目：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".

注意：s.substr(i,j) 作用，从第i个位置开始，取出长度为j的子串。

方法一：递归（内存溢出）

class Solution {
public:
    bool wordBreak(string s, unordered_set<string>& wordDict) {
        if(wordDict.find(s)!=wordDict.end())
            return true;
        for(int i=0;i<s.size();i++)
        {
            if((wordDict.find(s.substr(0,i))!=wordDict.end())&&(wordBreak(s.substr(i+1,s.size()-1),wordDict)==true))
                return true;
        }
        return false;
    }
};

Submission Result: Time Limit Exceeded

Last executed input:

"aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaab", ["a","aa","aaa","aaaa","aaaaa","aaaaaa","aaaaaaa","aaaaaaaa","aaaaaaaaa","aaaaaaaaaa"]