【陪伴式刷题】Day 41｜动态规划｜139.单词拆分（Word Break）

刷题顺序按照代码随想录建议

题目描述

英文版描述

Given a string s and a dictionary of strings wordDict, return true if s can be segmented into a space-separated sequence of one or more dictionary words.

Note that the same word in the dictionary may be reused multiple times in the segmentation.

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

Constraints:

1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s and wordDict[i] consist of only lowercase English letters.
All the strings of wordDict are unique.

英文版地址

leetcode.com/problems/wo...

中文版描述

给你一个字符串 s 和一个字符串列表 wordDict 作为字典。请你判断是否可以利用字典中出现的单词拼接出 s 。

注意： 不要求字典中出现的单词全部都使用，并且字典中的单词可以重复使用。

示例 1：

输入: s = "leetcode", wordDict = ["leet", "code"] 输出: true 解释: 返回 true 因为 "leetcode" 可以由 "leet" 和 "code" 拼接成。

示例 2：

arduino 复制代码

输入: s = "applepenapple", wordDict = ["apple", "pen"]
输出: true
解释: 返回 true 因为 "applepenapple" 可以由 "apple" "pen" "apple" 拼接成。
     注意，你可以重复使用字典中的单词。

示例 3：

输入: s = "catsandog", wordDict = ["cats", "dog", "sand", "and", "cat"] 输出: false

提示：

1 <= s.length <= 300
1 <= wordDict.length <= 1000
1 <= wordDict[i].length <= 20
s 和 wordDict[i] 仅由小写英文字母组成
wordDict 中的所有字符串 互不相同

中文版地址

leetcode.cn/problems/wo...

解题方法

java 复制代码

class Solution {
    public boolean wordBreak(String s, List<String> wordDict) {
        boolean[] dp = new boolean[s.length() + 1];
        dp[0] = true;

        for (int i = 1; i <= s.length(); i++) {
            for (String word : wordDict) {
                int len = word.length();
                if (i >= len && dp[i - len] && word.equals(s.substring(i - len, i))) {
                    dp[i] = true;
                    break;
                }
            }
        }

        return dp[s.length()];
    }
}

复杂度分析

时间复杂度：O(n^2)，其中 n 数组长度
空间复杂度：O(n)