leetcode 139. Word Break [dp]

最新推荐文章于 2025-04-13 18:39:57 发布

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最新推荐文章于 2025-04-13 18:39:57 发布

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分类专栏： leetcode 文章标签： dp 字符串

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本文探讨了如何判断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。通过使用动态规划方法，我们能够有效地解决这个问题，即使字典中的单词可以重复使用。

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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.

class Solution {
public:
    bool wordBreak(string s, vector<string>& wordDict) {
        // dp[i]表示字符串s[:i]能否拆分成符合要求的子字符串。我们可以看出，如果s[j:i]在给定的字符串组中，且dp[j]为True（即字符串s[:j]能够拆分成符合要求的子字符串）
        bool dp[s.size() + 1] = {false};
        dp[0] = true;
        for(int i = 1; i <= s.size(); i++){
            for(int j = i - 1; j >= 0; j--){
                string t = s.substr(j, i - j);
                if(dp[j] && find(wordDict.begin(), wordDict.end(), t) != wordDict.end()){
                    dp[i] = true;
                    break;
                }
            }
        }
        return dp[s.size()];
    }
};