LeetCode | 139. Word Break（DP）

最新推荐文章于 2025-04-13 18:39:57 发布

原创最新推荐文章于 2025-04-13 18:39:57 发布 · 137 阅读

0 ·

CC 4.0 BY-SA版权

LeetCode 专栏收录该内容

59 篇文章

订阅专栏

本文探讨了一道关于字符串能否通过字典中单词进行分割的问题，采用动态规划算法解决。通过实例解析，阐述了如何判断一个非空字符串是否能被拆分为字典中一个或多个单词的序列，并详细解释了算法实现过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Pick One

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

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        //所以这个和之前那个基本都要嵌套查找一遍
        //一点一点的往后顺
        boolean[] ans=new boolean[s.length()+1];
        Arrays.fill(ans,false);
        ans[0]=true;
        for(int i=1;i<ans.length;i++){
            for(int j=0;j<i;j++){
                //substring从0开始数，不包含结束位置
                if(ans[j]&&wordDict.contains(s.substring(j,i))){
                    //ans中表示的结果是从1开始数的，所以两个i正好是相对应的
                    ans[i]=true;
                }
            }
        }
        return ans[s.length()];
    }
}

题目是按照dp标签来刷的，所以导致想法有些局限了，总是想着要开辟一片存储空间来存放相应的内容，其实不然。所以说，对于dp的理解还不够深入，就像之前做的那道找出和为所有数一半的子集那道题目，一样是被局限住了

https://blog.youkuaiyun.com/xueying_2017/article/details/82876342